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You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
n = len(A)
if n == 1:
return 1
dp_even, dp_odd = [False] * (n + 1), [False] * (n + 1)
even_jumps, odd_jumps = [-1] * n, [-1] * n
asc_sorted = sorted((val, i) for i, val in enumerate(A))
desc_sorted = sorted((-val, i) for i, val in enumerate(A))
stack = deque()
for val, i in asc_sorted:
while stack and stack[-1] < i:
odd_jumps[stack.pop()] = i
stack.append(i)
stack = deque()
for val, i in desc_sorted:
while stack and stack[-1] < i:
even_jumps[stack.pop()] = i
stack.append(i)
dp_even[n - 1] = dp_odd[n - 1] = True
for i in reversed(range(n - 1)):
dp_even[i] = dp_odd[even_jumps[i]]
dp_odd[i] = dp_even[odd_jumps[i]]
return dp_odd.count(True) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR NUMBER VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: "List[int]") -> "int":
sorted_indexes = sorted(list(range(len(A))), key=lambda i: A[i])
oddnext = self.makeStack(sorted_indexes)
sorted_indexes.sort(key=lambda i: A[i], reverse=True)
evennext = self.makeStack(sorted_indexes)
odd = [False] * len(A)
even = [False] * len(A)
odd[-1] = even[-1] = True
for i in range(len(A) - 1, -1, -1):
if oddnext[i] is not None:
odd[i] = even[oddnext[i]]
if evennext[i] is not None:
even[i] = odd[evennext[i]]
return sum(odd)
def makeStack(self, sorted_indexes):
result = [None] * len(sorted_indexes)
stack = []
for i in sorted_indexes:
while stack and i > stack[-1]:
result[stack.pop()] = i
stack.append(i)
return result | CLASS_DEF FUNC_DEF STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NONE ASSIGN VAR VAR VAR VAR VAR IF VAR VAR NONE ASSIGN VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR STRING FUNC_DEF ASSIGN VAR BIN_OP LIST NONE FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR WHILE VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
increasing = [(v, i) for i, v in enumerate(A)]
increasing.sort()
increasing = [item[1] for item in increasing]
decreasing = [(v, i) for i, v in enumerate(A)]
decreasing.sort(key=lambda item: item[0] * -1)
decreasing = [item[1] for item in decreasing]
odds_next = make(increasing)
evens_next = make(decreasing)
odds = [(False) for i in range(len(A))]
odds[len(A) - 1] = True
evens = [(False) for i in range(len(A))]
evens[len(A) - 1] = True
results = [len(A) - 1]
for i in reversed(list(range(len(A) - 1))):
if odds_next[i] > i and evens[odds_next[i]]:
odds[i] = True
results.append(i)
if evens_next[i] > i and odds[evens_next[i]]:
evens[i] = True
print(results)
return len(results)
def make(arr):
output = [i for i in range(len(arr))]
stack = []
for i in range(len(arr)):
value = arr[i]
while stack and value > stack[-1]:
index = stack.pop()
output[index] = value
stack.append(value)
return output | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR LIST BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR WHILE VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
def helper(A):
ans = [0] * len(A)
stack = []
for a, i in A:
while stack and stack[-1] < i:
ans[stack.pop()] = i
stack.append(i)
return ans
odd = helper(sorted([a, i] for i, a in enumerate(A)))
even = helper(sorted([-a, i] for i, a in enumerate(A)))
l = len(A)
oddjump, evenjump = [0] * l, [0] * l
oddjump[-1] = evenjump[-1] = 1
for i in range(l - 1)[::-1]:
oddjump[i] = evenjump[odd[i]]
evenjump[i] = oddjump[even[i]]
return sum(oddjump) | CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | def get_nsge_ngse(a):
ngse_stack = []
nsge_stack = []
ngse = [-1] * len(a)
nsge = [-1] * len(a)
for i, item in sorted(enumerate(a), key=lambda x: (x[1], x[0])):
while nsge_stack and i > nsge_stack[-1]:
nsge[nsge_stack.pop()] = i
nsge_stack.append(i)
for i, item in sorted(enumerate(a), key=lambda x: (-1 * x[1], x[0])):
while ngse_stack and i > ngse_stack[-1]:
ngse[ngse_stack.pop()] = i
ngse_stack.append(i)
return nsge, ngse
def get_ways(a):
odd, even = get_nsge_ngse(a)
print(odd, even)
dp = [set() for x in a]
dp[-1].add("even")
dp[-1].add("odd")
count = 1
for i in range(len(a) - 2, -1, -1):
if odd[i] != -1 and "even" in dp[odd[i]]:
dp[i].add("odd")
if even[i] != -1 and "odd" in dp[even[i]]:
dp[i].add("even")
if "odd" in dp[i]:
count += 1
return count
class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
return get_ways(A) | FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER WHILE VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR NUMBER VAR NUMBER WHILE VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER STRING VAR VAR VAR EXPR FUNC_CALL VAR VAR STRING IF VAR VAR NUMBER STRING VAR VAR VAR EXPR FUNC_CALL VAR VAR STRING IF STRING VAR VAR VAR NUMBER RETURN VAR CLASS_DEF FUNC_DEF VAR VAR RETURN FUNC_CALL VAR VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
curr_list = []
smaller = []
larger = []
def find_insertion(ele):
left = 0
right = len(curr_list)
while left < right:
middle = (left + right) // 2
if curr_list[middle] < ele:
left = middle + 1
else:
right = middle
return left
for i in range(1, len(A) + 1):
index = find_insertion([A[-i], len(A) - i])
if index < len(curr_list) and curr_list[index][0] == A[-i]:
smaller.append(curr_list[index][1])
larger.append(curr_list[index][1])
curr_list[index][1] = len(A) - i
else:
if index > 0:
smaller.append(curr_list[index - 1][1])
else:
smaller.append(None)
if index < len(curr_list):
larger.append(curr_list[index][1])
else:
larger.append(None)
curr_list.insert(index, [A[-i], len(A) - i])
smaller.reverse()
larger.reverse()
can_reach_odd = [False] * len(A)
can_reach_even = [False] * len(A)
can_reach_odd[-1] = True
can_reach_even[-1] = True
for i in range(2, len(A) + 1):
if larger[-i] != None:
can_reach_odd[-i] = can_reach_even[larger[-i]]
if smaller[-i] != None:
can_reach_even[-i] = can_reach_odd[smaller[-i]]
solution = 0
for i in can_reach_odd:
if i:
solution += 1
return solution | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR LIST VAR VAR BIN_OP FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR NONE IF VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NONE EXPR FUNC_CALL VAR VAR LIST VAR VAR BIN_OP FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR NONE ASSIGN VAR VAR VAR VAR VAR IF VAR VAR NONE ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER RETURN VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
if not A:
return 0
n = len(A)
next_pos = [([-1] * 2) for _ in range(n)]
stack = []
idxes = sorted([i for i in range(n)], key=lambda i: (A[i], i))
for idx in idxes:
while stack and stack[-1] < idx:
next_pos[stack[-1]][1] = idx
stack.pop()
stack.append(idx)
stack = []
idxes = sorted([i for i in range(n)], key=lambda i: (-A[i], i))
for idx in idxes:
while stack and stack[-1] < idx:
next_pos[stack[-1]][0] = idx
stack.pop()
stack.append(idx)
dp = [([False] * 2) for _ in range(n)]
ret = 0
for i in range(n - 1, -1, -1):
if i == n - 1:
dp[i][0] = True
dp[i][1] = True
else:
odd_next = next_pos[i][1]
if odd_next == -1:
dp[i][1] = False
else:
dp[i][1] = dp[odd_next][0]
even_next = next_pos[i][0]
if even_next == -1:
dp[i][0] = False
else:
dp[i][0] = dp[even_next][1]
if dp[i][1] == True:
ret += 1
return ret | CLASS_DEF FUNC_DEF VAR VAR IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR VAR NUMBER IF VAR VAR NUMBER NUMBER VAR NUMBER RETURN VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
n, seen = len(A), {}
next_higher, next_lower = [0] * n, [0] * n
stack = []
for a, i in sorted([a, i] for i, a in enumerate(A)):
while stack and stack[-1] < i:
next_higher[stack.pop()] = i
stack.append(i)
stack = []
for a, i in sorted([-a, i] for i, a in enumerate(A)):
while stack and stack[-1] < i:
next_lower[stack.pop()] = i
stack.append(i)
def is_possible(ind, jump):
if ind == n - 1:
return True
if (ind, jump) in seen:
return seen[ind, jump]
temp = False
if jump % 2 == 0:
if next_lower[ind]:
temp = is_possible(next_lower[ind], jump + 1)
elif next_higher[ind]:
temp = is_possible(next_higher[ind], jump + 1)
seen[ind, jump] = temp
return temp
return sum(1 for i in range(n - 1, -1, -1) if is_possible(i, 1)) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR DICT ASSIGN VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FUNC_DEF IF VAR BIN_OP VAR NUMBER RETURN NUMBER IF VAR VAR VAR RETURN VAR VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR RETURN VAR RETURN FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FUNC_CALL VAR VAR NUMBER VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
if not A:
return 0
odd_dict = dict()
even_dict = dict()
n = len(A)
A_inc = sorted([(A[i], i) for i in range(len(A))])
A_dec = sorted([(-A[i], i) for i in range(len(A))])
odd_dict[A_inc[-1][1]] = -1
even_dict[A_dec[-1][1]] = -1
for i in range(n - 1):
j = i + 1
while j <= n - 1 and A_inc[j][1] < A_inc[i][1]:
j += 1
if j == n:
odd_dict[A_inc[i][1]] = -1
else:
odd_dict[A_inc[i][1]] = A_inc[j][1]
j = i + 1
while j <= n - 1 and A_dec[j][1] < A_dec[i][1]:
j += 1
if j == n:
even_dict[A_dec[i][1]] = -1
else:
even_dict[A_dec[i][1]] = A_dec[j][1]
print(odd_dict)
print(even_dict)
cnt = 1
for i in range(n - 1):
j = i
odd_flag = True
while j != n - 1 and j != -1:
if odd_flag:
j = odd_dict[j]
else:
j = even_dict[j]
odd_flag = not odd_flag
if j == n - 1:
cnt += 1
return cnt | CLASS_DEF FUNC_DEF VAR VAR IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER VAR NUMBER RETURN VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
n = len(A)
more = {}
less = {}
stack = []
B = sorted(range(n), key=lambda i: A[i])
for i in B:
while stack and stack[-1] < i:
more[stack.pop()] = i
stack.append(i)
B = sorted(range(n), key=lambda i: -A[i])
for i in B:
while stack and stack[-1] < i:
less[stack.pop()] = i
stack.append(i)
count = 0
for i in range(n):
jumps = 1
j = i
while j < n - 1:
if jumps % 2:
j = more[j] if j in more else None
else:
j = less[j] if j in less else None
if j is None:
break
jumps += 1
if j is not None:
count += 1
return count | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR DICT ASSIGN VAR DICT ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NONE ASSIGN VAR VAR VAR VAR VAR NONE IF VAR NONE VAR NUMBER IF VAR NONE VAR NUMBER RETURN VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
n = len(A)
next_large = [None] * n
next_small = [None] * n
arr1 = sorted([a, i] for i, a in enumerate(A))
stack = []
for a, i in arr1:
while stack and stack[-1] < i:
idx = stack.pop()
next_large[idx] = i
stack.append(i)
arr2 = sorted([-a, i] for i, a in enumerate(A))
stack = []
for a, i in arr2:
while stack and stack[-1] < i:
idx = stack.pop()
next_small[idx] = i
stack.append(i)
dp = [[None for _ in range(2)] for _ in range(n)]
def dfs(i, is_odd):
if i == n - 1:
return True
if dp[i][is_odd] is not None:
return dp[i][is_odd]
idx = next_large[i] if is_odd else next_small[i]
if idx is None:
dp[i][is_odd] = False
else:
dp[i][is_odd] = dfs(idx, is_odd ^ 1)
return dp[i][is_odd]
res = 0
for i in range(n):
if dfs(i, 1):
res += 1
return res | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NONE VAR ASSIGN VAR BIN_OP LIST NONE VAR ASSIGN VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR VAR WHILE VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NONE VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR BIN_OP VAR NUMBER RETURN NUMBER IF VAR VAR VAR NONE RETURN VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR IF VAR NONE ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR NUMBER RETURN VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | EVEN = 0
ODD = 1
class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
A_sorted = []
for i, a in enumerate(A):
item = a, i
A_sorted.append(item)
A_sorted.sort()
odd_jmp_dct = {}
for i, item in enumerate(A_sorted):
a, idx = item
if idx == len(A) - 1:
continue
j = i + 1
if j >= len(A_sorted):
continue
next_a, next_idx = A_sorted[j]
while j < len(A_sorted) - 1 and next_idx < idx:
j += 1
next_a, next_idx = A_sorted[j]
if next_idx > idx:
try:
odd_jmp_dct[next_idx].add(idx)
except KeyError:
odd_jmp_dct[next_idx] = set([idx])
A_sorted.sort(key=lambda x: (-x[0], x[1]))
even_jmp_dct = {}
for i, item in enumerate(A_sorted):
a, idx = item
if idx == len(A) - 1:
continue
j = i + 1
if j >= len(A_sorted):
continue
next_a, next_idx = A_sorted[j]
while j < len(A_sorted) - 1 and next_idx < idx:
j += 1
next_a, next_idx = A_sorted[j]
if next_idx > idx:
try:
even_jmp_dct[next_idx].add(idx)
except KeyError:
even_jmp_dct[next_idx] = set([idx])
q = collections.deque()
item1 = len(A) - 1, EVEN
item2 = len(A) - 1, ODD
q.append(item1)
q.append(item2)
ans = set([len(A) - 1])
while q:
i, jump_type = q.popleft()
if jump_type == ODD:
if i not in odd_jmp_dct:
continue
for new_i in odd_jmp_dct[i]:
new_item = new_i, EVEN
ans.add(new_i)
q.append(new_item)
else:
if i not in even_jmp_dct:
continue
for new_i in even_jmp_dct[i]:
new_item = new_i, ODD
q.append(new_item)
return len(ans) | ASSIGN VAR NUMBER ASSIGN VAR NUMBER CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR LIST VAR EXPR FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR DICT FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR LIST VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR LIST BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR ASSIGN VAR VAR FUNC_CALL VAR IF VAR VAR IF VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
odd_jumps = {}
stack = []
for num, i in sorted([[num, i] for i, num in enumerate(A)], reverse=True):
while stack and stack[-1][1] < i:
stack.pop()
if not stack:
odd_jumps[i] = None
else:
odd_jumps[i] = stack[-1][1]
stack.append([num, i])
even_jumps = {}
stack = []
for num, i in sorted([[num, -i] for i, num in enumerate(A)]):
i = -i
while stack and stack[-1][1] < i:
stack.pop()
if not stack:
even_jumps[i] = None
else:
even_jumps[i] = stack[-1][1]
stack.append([num, i])
n = len(A)
@lru_cache(None)
def dfs(i, is_odd):
if i == n - 1:
return True
if is_odd and odd_jumps[i]:
return dfs(odd_jumps[i], False)
elif not is_odd and even_jumps[i]:
return dfs(even_jumps[i], True)
else:
return False
res = 0
for i in range(n):
res += 1 if dfs(i, True) else 0
return res | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER WHILE VAR VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR IF VAR ASSIGN VAR VAR NONE ASSIGN VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR LIST VAR VAR ASSIGN VAR DICT ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR IF VAR ASSIGN VAR VAR NONE ASSIGN VAR VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR LIST VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR BIN_OP VAR NUMBER RETURN NUMBER IF VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR NUMBER RETURN NUMBER FUNC_CALL VAR NONE ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER NUMBER RETURN VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, A: List[int]) -> int:
n = len(A)
s1 = []
s2 = []
stack = []
e = enumerate(A)
order = sorted(e, key=lambda x: (x[1], x[0]))
for i in reversed(list(range(n))):
while stack and stack[-1] < order[i][0]:
stack.pop()
order[i] = (order[i][0], stack[-1]) if stack else (order[i][0], -1)
stack.append(order[i][0])
arr_odd = [-1] * n
for i in range(n):
arr_odd[order[i][0]] = order[i][1]
stack = []
e = enumerate(A)
order = sorted(e, key=lambda x: x[1], reverse=True)
for i in reversed(list(range(n))):
while stack and stack[-1] < order[i][0]:
stack.pop()
order[i] = (order[i][0], stack[-1]) if stack else (order[i][0], -1)
stack.append(order[i][0])
arr_even = [-1] * n
for i in range(n):
arr_even[order[i][0]] = order[i][1]
DP = [([False] * 2) for _ in range(n)]
DP[n - 1][0] = True
DP[n - 1][1] = True
for i in reversed(list(range(n - 1))):
if arr_odd[i] == n - 1:
DP[i][0] = True
elif arr_odd[i] != -1:
DP[i][0] = DP[arr_odd[i]][1]
if arr_even[i] == n - 1:
DP[i][1] = True
elif arr_even[i] != -1:
DP[i][1] = DP[arr_even[i]][0]
count = 0
for i in range(n):
if DP[i][0] == True:
count += 1
return count | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR WHILE VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR WHILE VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR VAR VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER NUMBER VAR NUMBER RETURN VAR VAR |
You are given an integer array A. From some starting index, you can make a series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, ...) jumps in the series are called even numbered jumps.
You may from index i jump forward to index j (with i < j) in the following way:
During odd numbered jumps (ie. jumps 1, 3, 5, ...), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
During even numbered jumps (ie. jumps 2, 4, 6, ...), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value. If there are multiple such indexes j, you can only jump to the smallest such index j.
(It may be the case that for some index i, there are no legal jumps.)
A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)
Return the number of good starting indexes.
Example 1:
Input: [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[0]), then we can't jump any more.
From starting index i = 1 and i = 2, we can jump to i = 3, then we can't jump any more.
From starting index i = 3, we can jump to i = 4, so we've reached the end.
From starting index i = 4, we've reached the end already.
In total, there are 2 different starting indexes (i = 3, i = 4) where we can reach the end with some number of jumps.
Example 2:
Input: [2,3,1,1,4]
Output: 3
Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd numbered), we first jump to i = 1 because A[1] is the smallest value in (A[1], A[2], A[3], A[4]) that is greater than or equal to A[0].
During our 2nd jump (even numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in (A[2], A[3], A[4]) that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3.
During our 3rd jump (odd numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in (A[3], A[4]) that is greater than or equal to A[2].
We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indexes (i = 1, i = 3, i = 4) where we can reach the end with some number of jumps.
Example 3:
Input: [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indexes 1, 2, and 4.
Note:
1 <= A.length <= 20000
0 <= A[i] < 100000 | class Solution:
def oddEvenJumps(self, a: List[int]) -> int:
b = list(sorted(list(range(len(a))), key=lambda i: a[i]))
c = list(sorted(list(range(len(a))), key=lambda i: -a[i]))
odd_next = {}
even_next = {}
i = 1
while i < len(b):
j = i
while j < len(b) - 1 and b[j] <= b[i - 1]:
j += 1
if b[j] > b[i - 1]:
odd_next[b[i - 1]] = b[j]
i += 1
i = 1
while i < len(c):
j = i
while j < len(c) - 1 and c[j] <= c[i - 1]:
j += 1
if c[j] > c[i - 1]:
even_next[c[i - 1]] = c[j]
i += 1
paths = 0
for start in list(odd_next.keys()):
index = start
can_go = True
odd = True
while can_go and index != len(a) - 1:
if odd:
if index in list(odd_next.keys()):
index = odd_next[index]
else:
can_go = False
elif index in list(even_next.keys()):
index = even_next[index]
else:
can_go = False
odd = not odd
if index == len(a) - 1:
paths += 1
if len(a) - 1 not in list(odd_next.keys()):
paths += 1
return paths | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR DICT ASSIGN VAR DICT ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR IF VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER IF VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER RETURN VAR VAR |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
m, n = len(dungeon), len(dungeon[0])
record = [([None] * n) for i in range(m)]
record[-1][-1] = max(1 - dungeon[-1][-1], 1)
for i in range(n - 1)[::-1]:
record[m - 1][i] = max(1, record[m - 1][i + 1] - dungeon[m - 1][i])
for j in range(m - 1)[::-1]:
record[j][n - 1] = max(1, record[j + 1][n - 1] - dungeon[j][n - 1])
for i in range(m - 1)[::-1]:
for j in range(n - 1)[::-1]:
mina = max(record[i + 1][j] - dungeon[i][j], 1)
minb = max(record[i][j + 1] - dungeon[i][j], 1)
record[i][j] = min(mina, minb)
return record[0][0] | CLASS_DEF FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NONE VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP NUMBER VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR NUMBER NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
if not dungeon or not dungeon[0]:
return 1
m = len(dungeon)
n = len(dungeon[0])
dp = [None] * n
for i in range(m - 1, -1, -1):
for j in range(n - 1, -1, -1):
tmp = []
if i + 1 < m:
tmp.append(dp[j])
if j + 1 < n:
tmp.append(dp[j + 1])
minLeft = 1
if tmp:
minLeft = min(tmp)
dp[j] = max(minLeft - dungeon[i][j], 1)
return dp[0] | CLASS_DEF FUNC_DEF IF VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NONE VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR LIST IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR NUMBER RETURN VAR NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
num_rows = len(dungeon)
num_cols = len(dungeon[0])
INF = float("inf")
for ri in range(num_rows - 1, -1, -1):
for ci in range(num_cols - 1, -1, -1):
if ri == num_rows - 1 and ci == num_cols - 1:
need = -dungeon[ri][ci] + 1
else:
down = INF if ri == num_rows - 1 else dungeon[ri + 1][ci]
right = INF if ci == num_cols - 1 else dungeon[ri][ci + 1]
need = min(down, right) - dungeon[ri][ci]
dungeon[ri][ci] = max(need, 1)
return dungeon[0][0] | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR NUMBER RETURN VAR NUMBER NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
best = [([0] * len(dungeon[0])) for x in range(0, len(dungeon))]
for row in range(-1, -len(dungeon) - 1, -1):
for col in range(-1, -len(dungeon[0]) - 1, -1):
needed = -dungeon[row][col]
nextSteps = []
if row < -1:
nextSteps.append(best[row + 1][col])
if col < -1:
nextSteps.append(best[row][col + 1])
if len(nextSteps) > 0:
needed += min(nextSteps)
needed = max(0, needed)
best[row][col] = needed
return best[0][0] + 1 | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR LIST IF VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR VAR RETURN BIN_OP VAR NUMBER NUMBER NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
m = len(dungeon)
n = len(dungeon[0])
dp = [[float("inf") for _ in range(n + 1)] for _ in range(m + 1)]
dp[m][n - 1] = 1
dp[m - 1][n] = 1
for i in range(m - 1, -1, -1):
for j in range(n - 1, -1, -1):
need = min(dp[i + 1][j], dp[i][j + 1]) - dungeon[i][j]
if need <= 0:
need = 1
dp[i][j] = need
return dp[0][0] | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR RETURN VAR NUMBER NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
n = len(dungeon[0])
need = [2**31] * (n - 1) + [1]
for row in dungeon[::-1]:
for j in range(n)[::-1]:
need[j] = max(min(need[j : j + 2]) - row[j], 1)
return need[0] | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP LIST BIN_OP NUMBER NUMBER BIN_OP VAR NUMBER LIST NUMBER FOR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER RETURN VAR NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
if not any(dungeon):
return 1
m = len(dungeon)
n = len(dungeon[0])
memo = {(m - 1, n - 1): 1}
def helper(i, j):
if (i, j) not in memo:
if i == m - 1:
ans = max(1, helper(i, j + 1) - dungeon[i][j + 1])
elif j == n - 1:
ans = max(1, helper(i + 1, j) - dungeon[i + 1][j])
else:
ans = min(
max(1, helper(i, j + 1) - dungeon[i][j + 1]),
max(1, helper(i + 1, j) - dungeon[i + 1][j]),
)
memo[i, j] = ans
return memo[i, j]
return max(1, helper(0, 0) - dungeon[0][0]) | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR DICT BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER FUNC_DEF IF VAR VAR VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR VAR RETURN VAR VAR VAR RETURN FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR NUMBER NUMBER VAR NUMBER NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
if not dungeon or not dungeon[0]:
return 1
dp = [[(0) for i in range(len(dungeon[0]))] for i in range(len(dungeon))]
dp[-1][-1] = 1 - dungeon[-1][-1]
row = len(dungeon)
column = len(dungeon[0])
for i in range(1, row):
dp[row - 1 - i][-1] = max(
1 - dungeon[row - 1 - i][-1], dp[row - i][-1] - dungeon[row - 1 - i][-1]
)
for i in range(1, column):
dp[-1][column - 1 - i] = max(
1 - dungeon[-1][column - 1 - i],
dp[-1][column - i] - dungeon[-1][column - 1 - i],
)
for i in range(1, row):
for j in range(1, column):
dp[row - 1 - i][column - 1 - j] = max(
1 - dungeon[row - 1 - i][column - 1 - j],
min(dp[row - i][column - 1 - j], dp[row - 1 - i][column - j])
- dungeon[row - 1 - i][column - 1 - j],
)
if dp[0][0] <= 0:
return 1
return dp[0][0] | CLASS_DEF FUNC_DEF IF VAR VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER NUMBER BIN_OP NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP VAR BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP NUMBER VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR NUMBER NUMBER NUMBER RETURN NUMBER RETURN VAR NUMBER NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
n = len(dungeon)
m = len(dungeon[0])
for i in range(n - 1, -1, -1):
for j in range(m - 1, -1, -1):
if i + 1 < n and j + 1 < m:
temp = max(dungeon[i + 1][j], dungeon[i][j + 1])
elif i + 1 < n:
temp = dungeon[i + 1][j]
elif j + 1 < m:
temp = dungeon[i][j + 1]
else:
temp = 0
print(temp)
dungeon[i][j] = dungeon[i][j] + temp
dungeon[i][j] = min(dungeon[i][j], 0)
return max(1, 1 - dungeon[0][0]) | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR NUMBER BIN_OP NUMBER VAR NUMBER NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
row = len(dungeon)
col = len(dungeon[0])
dp = [[(0) for c in range(col)] for r in range(row)]
if dungeon[-1][-1] <= 0:
dp[-1][-1] = -dungeon[-1][-1] + 1
else:
dp[-1][-1] = 1
for r in range(row - 2, -1, -1):
dp[r][-1] = dp[r + 1][-1] - dungeon[r][-1]
if dp[r][-1] <= 0:
dp[r][-1] = 1
for c in range(col - 2, -1, -1):
dp[-1][c] = dp[-1][c + 1] - dungeon[-1][c]
if dp[-1][c] <= 0:
dp[-1][c] = 1
for r in range(row - 2, -1, -1):
for c in range(col - 2, -1, -1):
dp[r][c] = min(dp[r + 1][c], dp[r][c + 1]) - dungeon[r][c]
if dp[r][c] <= 0:
dp[r][c] = 1
return dp[0][0] | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER IF VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR NUMBER NUMBER |
table.dungeon, .dungeon th, .dungeon td {
border:3px solid black;
}
.dungeon th, .dungeon td {
text-align: center;
height: 70px;
width: 70px;
}
The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.
The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.
Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).
In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.
Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.
For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.
-2 (K)
-3
3
-5
-10
1
10
30
-5 (P)
Note:
The knight's health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. | class Solution:
def calculateMinimumHP(self, dungeon):
M, N = len(dungeon), len(dungeon[0])
dp = [([0] * N) for _ in range(M)]
for i in reversed(list(range(M))):
for j in reversed(list(range(N))):
v = dungeon[i][j]
if i == M - 1 and j == N - 1:
dp[i][j] = 1 - v
elif i == M - 1:
dp[i][j] = dp[i][j + 1] - v
elif j == N - 1:
dp[i][j] = dp[i + 1][j] - v
else:
dp[i][j] = min(dp[i + 1][j], dp[i][j + 1]) - v
dp[i][j] = max(1, dp[i][j])
return dp[0][0] | CLASS_DEF FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP NUMBER VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR FUNC_CALL VAR NUMBER VAR VAR VAR RETURN VAR NUMBER NUMBER |
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example:
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4 | class Solution:
def maximalSquare(self, matrix):
if len(matrix) == 0 or len(matrix[0]) == 0:
return 0
dp = [[(0) for i in range(len(matrix[0]))] for i in range(len(matrix))]
largest = 0
for i in range(len(matrix)):
dp[i][0] = int(matrix[i][0])
largest = max(largest, dp[i][0])
for j in range(len(matrix[0])):
dp[0][j] = int(matrix[0][j])
largest = max(largest, dp[0][j])
for i in range(1, len(matrix)):
for j in range(1, len(matrix[0])):
if matrix[i][j] == "1":
if (
dp[i - 1][j] >= dp[i - 1][j - 1]
and dp[i][j - 1] >= dp[i - 1][j - 1]
):
dp[i][j] = dp[i - 1][j - 1] + 1
else:
dp[i][j] = min(int(dp[i - 1][j]), int(dp[i][j - 1])) + 1
else:
dp[i][j] = 0
largest = max(largest, dp[i][j])
return largest * largest | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER RETURN NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR STRING IF VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN BIN_OP VAR VAR |
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example:
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4 | class Solution:
def maximalSquare(self, matrix):
if len(matrix) == 0 or len(matrix[0]) == 0:
return 0
n, m = len(matrix), len(matrix[0])
dp = [[(0) for i in range(m)] for j in range(n)]
for i in range(n):
dp[i][0] = int(matrix[i][0])
for j in range(m):
dp[0][j] = int(matrix[0][j])
for i in range(1, n):
for j in range(1, m):
if matrix[i][j] == "1":
dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
ans = 0
for i in range(n):
ans = max(ans, max(dp[i]))
return ans**2 | CLASS_DEF FUNC_DEF IF FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR STRING ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR RETURN BIN_OP VAR NUMBER |
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example:
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4 | class Solution:
def maximalSquare(self, matrix):
m, n = len(matrix), len(matrix[0]) if matrix else 0
dp = [0] * n
best = 0
for r in range(m):
ndp = [0] * n
for c in range(n):
if matrix[r][c] == "1":
ndp[c] = min(dp[c - 1], dp[c], ndp[c - 1]) + 1 if r and c else 1
if ndp[c] > best:
best = ndp[c]
dp = ndp
return best**2 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR STRING ASSIGN VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR RETURN BIN_OP VAR NUMBER |
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example:
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4 | class Solution:
def maximalSquare(self, matrix):
if not matrix or len(matrix) == 0:
return 0
m = len(matrix)
n = len(matrix[0])
dp = [[(0) for i in range(n)] for j in range(m)]
dp[0] = list(map(lambda x: int(x), matrix[0]))
maxLength = 1 if 1 in dp[0] else 0
for i in range(1, m):
dp[i][0] = int(matrix[i][0])
if dp[i][0] == 1:
maxLength = 1
for i in range(1, m):
for j in range(1, n):
if matrix[i][j] == "0":
dp[i][j] = 0
else:
dp[i][j] = 1 + min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1])
maxLength = max(maxLength, dp[i][j])
return maxLength**2 | CLASS_DEF FUNC_DEF IF VAR FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR STRING ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN BIN_OP VAR NUMBER |
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example:
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4 | class Solution:
def maximalSquare(self, matrix):
if not matrix:
return 0
m, n = len(matrix), len(matrix[0])
dp = [int(matrix[i][0]) for i in range(m)]
vmax = max(dp)
pre = 0
for j in range(1, n):
pre, dp[0] = int(matrix[0][j - 1]), int(matrix[0][j])
for i in range(1, m):
cur = dp[i]
dp[i] = 0 if matrix[i][j] == "0" else min(dp[i - 1], dp[i], pre) + 1
pre = cur
vmax = max(vmax, max(dp))
return vmax**2 | CLASS_DEF FUNC_DEF IF VAR RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR STRING NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN BIN_OP VAR NUMBER |
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example:
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4 | class Solution:
def maximalSquare(self, matrix):
g = 0
li = [[(0) for a in x] for x in matrix]
for i in range(0, len(matrix)):
for j in range(0, len(matrix[0])):
c = int(matrix[i][j])
if c == 0:
li[i][j] = 0
continue
if i == 0 or j == 0:
li[i][j] = c
if c > g:
g = c
continue
m = min(li[i - 1][j], li[i][j - 1])
if li[i - 1][j - 1] <= m:
li[i][j] = 1 + li[i - 1][j - 1]
else:
li[i][j] = 1 + m
if li[i][j] > g:
g = li[i][j]
return g**2 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP NUMBER VAR IF VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR RETURN BIN_OP VAR NUMBER |
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example:
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4 | class Solution:
def maximalSquare(self, matrix):
if not matrix:
return 0
rows, cols = len(matrix), len(matrix[0])
dp = [0] * cols
maxLen = 1 if sum(dp) > 0 else 0
for i in range(0, rows):
for j in range(0, cols):
if matrix[i][j] == "1":
if j == 0:
dp[j] = int(matrix[i][j])
else:
k = min(dp[j], dp[j - 1])
dp[j] = k + 1 if matrix[i - k][j - k] == "1" else k
maxLen = max(maxLen, dp[j])
else:
dp[j] = 0
return maxLen * maxLen | CLASS_DEF FUNC_DEF IF VAR RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR STRING IF VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR STRING BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR NUMBER RETURN BIN_OP VAR VAR |
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example:
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4 | class Solution:
def maximalSquare(self, matrix):
if not matrix:
return 0
rows, cols = len(matrix), len(matrix[0])
dp = list(map(int, matrix[0][:]))
maxLen = 1 if sum(dp) > 0 else 0
for i in range(1, rows):
tmp = dp[0]
dp[0] = int(matrix[i][0])
maxLen = max(maxLen, dp[0])
pre = tmp
for j in range(1, cols):
tmp = dp[j]
if matrix[i][j] == "1":
dp[j] = min(dp[j], dp[j - 1], pre) + 1
maxLen = max(maxLen, dp[j])
else:
dp[j] = 0
pre = tmp
return maxLen * maxLen | CLASS_DEF FUNC_DEF IF VAR RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR IF VAR VAR VAR STRING ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR RETURN BIN_OP VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | def main():
k1, k2, k3 = map(int, input().split())
n = k1 + k2 + k3
arr = [(0) for _ in range(n)]
a = list(map(int, input().split()))
for i in a:
arr[i - 1] = 1
a = list(map(int, input().split()))
for i in a:
arr[i - 1] = 2
a = list(map(int, input().split()))
for i in a:
arr[i - 1] = 3
dp = [0, 0, 0]
for i in arr:
if i == 1:
dp[1] += 1
dp[2] += 1
if i == 2:
dp[1] = min(dp[0], dp[1])
dp[0] += 1
dp[2] += 1
if i == 3:
dp[2] = min(dp[0], dp[1], dp[2])
dp[0] += 1
dp[1] += 1
print(min(dp))
def __starting_point():
main()
__starting_point() | FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER FOR VAR VAR IF VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | def solve(a, b, c):
n = len(a) + len(b) + len(c)
arr = [0] * (n + 1)
for i in a:
arr[i] = 0
for i in b:
arr[i] = 1
for i in c:
arr[i] = 2
dp = [([n + 10] * (n + 1)) for _ in range(3)]
for i in range(3):
dp[i][0] = 0
for j in range(1, n + 1):
dp[i][j] = dp[i][j - 1] + (arr[j] != i)
if i:
dp[i][j] = min(dp[i - 1][j], dp[i][j])
return dp[2][n]
def __starting_point():
k1, k2, k3 = map(int, input().split(" "))
n = k1 + k2 + k3
a = list(map(int, input().split(" ")))
b = list(map(int, input().split(" ")))
c = list(map(int, input().split(" ")))
assert n == len(a) + len(b) + len(c)
arr = [(0) for _ in range(n)]
for i in a:
arr[i - 1] = 1
for i in b:
arr[i - 1] = 2
for i in c:
arr[i - 1] = 3
dp = [0, 0, 0]
print(solve(a, b, c))
__starting_point() | FUNC_DEF ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP LIST BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR RETURN VAR NUMBER VAR FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | import sys
k1, k2, k3 = [int(i) for i in sys.stdin.readline().split()]
k = [[], [], [], []]
for j in range(1, 4):
k[j] = [int(i) for i in sys.stdin.readline().split()]
n = k1 + k2 + k3
k_4 = []
for f in [2, 3]:
k_4.extend((jlk, f) for jlk in k[f])
k_4.sort(reverse=True)
ndd = [0] * (k2 + k3 + 1)
for j in range(1, k2 + k3 + 1):
if k_4[j - 1][1] == 3:
ndd[j] = ndd[j - 1]
else:
ndd[j] = ndd[j - 1] + 1
fj = []
for j in range(k2 + k3 + 1):
fj.append(k3 - j + 2 * ndd[j])
mnfj_i = []
mn = fj[0]
for i in range(k2 + k3 + 1):
mn = min(mn, fj[i])
mnfj_i.append(mn)
k3s = set(k[3])
cnt_lea3 = [0] * (n + 1)
for i in range(1, n + 1):
if i in k3s:
cnt_lea3[i] = cnt_lea3[i - 1] + 1
else:
cnt_lea3[i] = cnt_lea3[i - 1]
k2s = set(k[2])
cnt_lea2 = [0] * (n + 1)
for i in range(1, n + 1):
if i in k2s:
cnt_lea2[i] = cnt_lea2[i - 1] + 1
else:
cnt_lea2[i] = cnt_lea2[i - 1]
k1s = set(k[1])
cnt_lea1 = [0] * (n + 1)
for i in range(1, n + 1):
if i in k1s:
cnt_lea1[i] = cnt_lea1[i - 1] + 1
else:
cnt_lea1[i] = cnt_lea1[i - 1]
ans = 5 * (k1 + k2 + k3)
for alpha in range(n + 1):
ub = k2 + k3 - cnt_lea3[alpha] - cnt_lea2[alpha]
ans = min(ans, mnfj_i[ub] - cnt_lea3[alpha] + k1 + alpha - 2 * cnt_lea1[alpha])
print(ans) | IMPORT ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST LIST LIST LIST LIST FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR LIST FOR VAR LIST NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP NUMBER VAR VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP NUMBER BIN_OP BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | numbers = list(map(int, input().split()))
k1 = numbers[0]
k2 = numbers[1]
k3 = numbers[2]
n = 3
suma = 0
for i, elem in enumerate(numbers):
suma = suma + elem
lst = []
for i in range(suma):
lst.append(0)
for i in range(3):
problems = list(map(int, input().split()))
for j in problems:
lst[j - 1] = i
p = []
for i in range(3):
p.append(-1)
vect = []
for i in range(suma + 1):
vect.append(1)
for i in range(suma):
for j in range(lst[i] + 1):
if p[j] >= 0:
vect[i] = max(vect[i], 1 + vect[p[j]])
x = lst[i]
p[lst[i]] = i
maxi = -1
for i in vect:
if maxi < i:
maxi = i
print(suma - maxi) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | from sys import stdin
input = stdin.readline
r = list(map(int, input().split()))
n = sum(r)
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
mp = [0] * -~n
for i in a:
mp[i] = 0
for i in b:
mp[i] = 1
for i in c:
mp[i] = 2
ans = sum(r[:2])
now = ans
sm0 = 0
sm2 = r[2]
for i in range(1, n + 1):
sm0 += mp[i] == 0
sm2 -= mp[i] == 2
if mp[i] == 2:
now += 1
elif mp[i] == 1:
now -= 1
else:
now = min(now, n - sm0 - sm2)
ans = min(ans, now)
print(ans) | ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR NUMBER VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR NUMBER IF VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | x, y, z = map(int, input().split())
d = [[], [], []]
d[0] = list(map(int, input().split()))
d[1] = list(map(int, input().split()))
d[2] = list(map(int, input().split()))
dicto = [dict(), dict(), dict()]
dp = [([float("inf")] * (x + y + z + 1)) for i in range(3)]
for i in range(3):
for j in d[i]:
dicto[i][j] = 1
dp[0][0] = 0
mini = 9999999999999
for i in range(1, x + y + z + 1):
mini = 999999999999
for j in range(3):
for k in range(j + 1):
mini = min(mini, dp[k][i - 1])
if i not in dicto[j]:
dp[j][i] = mini + 1
else:
dp[j][i] = mini
for i in range(3):
mini = min(dp[0][x + y + z], dp[1][x + y + z], dp[2][x + y + z])
print(mini) | ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST LIST LIST LIST ASSIGN VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST FUNC_CALL VAR STRING BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | k1, k2, k3 = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
def ceil(a, L, R, pivot):
while L + 1 < R:
m = (L + R) // 2
if a[m] > pivot:
R = m
else:
L = m
return R
def LIS(a, n):
tails = [(0) for _ in range(n + 1)]
empty = 0
tails[0] = a[0]
empty = 1
for i in range(1, n):
if a[i] < tails[0]:
tails[0] = a[i]
elif a[i] >= tails[empty - 1]:
tails[empty] = a[i]
empty += 1
else:
tails[ceil(tails, -1, empty - 1, a[i])] = a[i]
return empty
n = k1 + k2 + k3
abc = [None] * n
for el in a:
abc[el - 1] = 1
for el in b:
abc[el - 1] = 2
for el in c:
abc[el - 1] = 3
print(n - LIS(abc, n)) | ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF WHILE BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR VAR RETURN VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP LIST NONE VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | k1, k2, k3 = map(int, input().split())
s = [set([int(o) for o in input().split()]) for i in range(3)]
n = k1 + k2 + k3
dp = [([float("inf")] * 3) for i in range(n + 1)]
dp[0][0] = 0
for i in range(1, n + 1):
for j in range(3):
for k in range(j, 3):
dp[i][k] = min(dp[i][k], dp[i - 1][j] + (i not in s[k]))
print(min(dp[n])) | ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP LIST FUNC_CALL VAR STRING NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | k1, k2, k3 = map(int, input().split())
x1 = list(map(int, input().split()))
x2 = list(map(int, input().split()))
x3 = list(map(int, input().split()))
y = [0] * (k1 + k2 + k3 + 1)
dp = [([0] * 3) for i in range(k1 + k2 + k3 + 1)]
for i in range(k1):
y[x1[i]] = 1
for i in range(k2):
y[x2[i]] = 2
for i in range(k3):
y[x3[i]] = 3
for i in range(1, k1 + k2 + k3 + 1):
dp[i][0] = dp[i - 1][0] + (y[i] != 1)
dp[i][1] = min(dp[i - 1][0], dp[i - 1][1]) + (y[i] != 2)
dp[i][2] = min(dp[i - 1][0], dp[i - 1][1], dp[i - 1][2]) + (y[i] != 3)
print(min(dp[k1 + k2 + k3])) | ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | k1, k2, k3 = map(int, input().split())
n = k1 + k2 + k3
a = [0] * n
input()
for x in map(int, input().split()):
a[x - 1] = 1
for x in map(int, input().split()):
a[x - 1] = 2
prev = [0, 0, 0]
for i, x in enumerate(a):
new = [0, 0, 0]
new[0] = prev[0] + (x != 0)
new[1] = min(prev[0], prev[1]) + (x != 1)
new[2] = min(prev) + (x != 2)
prev = new
print(min(prev)) | ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR LIST NUMBER NUMBER NUMBER FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER ASSIGN VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | k1, k2, k3 = map(int, input().split())
l1 = list(map(int, input().split()))
l2 = list(map(int, input().split()))
l3 = list(map(int, input().split()))
n = k1 + k2 + k3
assign = [0] * n
for i in l1:
assign[i - 1] = 1
for i in l2:
assign[i - 1] = 2
for i in l3:
assign[i - 1] = 3
ilenie3wprawo = [0] * n
ilenie1wlewo = [0] * n
ilenie1wlewo[n - 1] = k2 + k3
i = n - 1
ilenie3wprawo[n - 1] = 1 if assign[n - 1] != 3 else 0
while i > 0:
i -= 1
ilenie3wprawo[i] = ilenie3wprawo[i + 1]
ilenie1wlewo[i] = ilenie1wlewo[i + 1]
if assign[i] != 3:
ilenie3wprawo[i] += 1
if assign[i + 1] != 1:
ilenie1wlewo[i] -= 1
odp = [0] * (n + 1)
odp[n] = k2 + k3
i = n
while i > 1:
i -= 1
if assign[i] == 2:
odp[i] = odp[i + 1] - 1
if assign[i] == 1:
odp[i] = odp[i + 1] + 1
if assign[i] == 3:
odp[i] = min(odp[i + 1], ilenie1wlewo[i - 1] + ilenie3wprawo[i])
if assign[0] == 2:
odp[0] = odp[1] - 1
if assign[0] == 1:
odp[0] = odp[1] + 1
if assign[0] == 3:
odp[0] = min(odp[i + 1], ilenie3wprawo[0])
print(min(odp)) | ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER NUMBER NUMBER WHILE VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR NUMBER VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR VAR WHILE VAR NUMBER VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER VAR VAR IF VAR NUMBER NUMBER ASSIGN VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR NUMBER NUMBER ASSIGN VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR NUMBER NUMBER ASSIGN VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | k1, k2, k3 = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
a = sorted(a)
b = sorted(b)
c = sorted(c)
n = k1 + k2 + k3
l = [0] * (n + 1)
r = [0] * (n + 1)
p = 0
for i in range(n + 1):
if p < len(a) and i == a[p]:
p += 1
l[i] += len(a) - p
p = 0
for i in range(n + 1):
if p < len(b) and i == b[p]:
p += 1
l[i] += p
p = 0
for i in range(n + 1):
if p < len(b) and i == b[p]:
p += 1
r[i] += len(b) - p
p = 0
for i in range(n + 1):
if p < len(c) and i == c[p]:
p += 1
r[i] += p
r_cum_min = [None] * (n + 1)
r_cum_min[-1] = r[n]
i = n - 1
while i >= 0:
r_cum_min[i] = min(r[i], r_cum_min[i + 1])
i -= 1
answer = 1000000000.0
for i in range(n + 1):
if l[i] + r_cum_min[i] < answer:
answer = l[i] + r_cum_min[i]
print(answer) | ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP LIST NONE BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | n = sum(list(map(int, input().split())))
A = [(0) for i in range(n)]
DP = [[(0) for i in range(3)] for i in range(n)]
for i in range(3):
for j in map(int, input().split()):
A[j - 1] = i
DP[0] = [(A[0] != i) for i in range(3)]
for i in range(1, n):
DP[i][0] = DP[i - 1][0] + (A[i] != 0)
DP[i][1] = min(DP[i - 1][0], DP[i - 1][1]) + (A[i] != 1)
DP[i][2] = min(DP[i - 1]) + (A[i] != 2)
print(min(DP[n - 1])) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER VAR NUMBER VAR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | import sys
input = sys.stdin.readline
K1, K2, K3 = map(int, input().split())
N = K1 + K2 + K3
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))
A.sort()
B.sort()
C.sort()
state = []
P = []
ind = 0
for b in B:
while ind < K3 and C[ind] < b:
state.append(1)
P.append(C[ind])
ind += 1
state.append(0)
P.append(b)
while ind < K3:
state.append(1)
P.append(C[ind])
ind += 1
dp = [[0, 0] for _ in range(K2 + K3 + 1)]
for i in reversed(range(K2 + K3)):
if state[i] == 0:
dp[i][0] = min(dp[i + 1][0], dp[i + 1][1])
dp[i][1] = dp[i + 1][1] + 1
else:
dp[i][0] = min(dp[i + 1][0], dp[i + 1][1]) + 1
dp[i][1] = dp[i + 1][1]
ans = 10**14
indp = 0
inda = 0
for n in range(N + 1):
while indp < K3 + K2 and P[indp] <= n:
indp += 1
while inda < K1 and A[inda] <= n:
inda += 1
ans = min(ans, indp + (K1 - inda) + min(dp[indp]))
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR WHILE VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR WHILE VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR LIST NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER WHILE VAR BIN_OP VAR VAR VAR VAR VAR VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | a, b, c = [int(i) for i in input().split()]
l = [(0) for i in range(a + b + c)]
arr1 = [int(i) for i in input().split()]
for i in arr1:
l[i - 1] = 1
arr2 = [int(i) for i in input().split()]
for i in arr2:
l[i - 1] = 2
arr3 = [int(i) for i in input().split()]
for i in arr3:
l[i - 1] = 3
l1 = [(0) for i in range(a + b + c + 1)]
for i in range(a + b + c):
if l[i] == 1:
l1[i + 1] = l1[i] - 1
elif l[i] == 2:
l1[i + 1] = l1[i] + 1
else:
l1[i + 1] = l1[i]
diff = [(0) for i in range(a + b + c + 1)]
for i in range(a + b + c - 1, -1, -1):
if l[i] == 2:
diff[i] = diff[i + 1] + 1
elif l[i] == 3:
diff[i] = diff[i + 1] - 1
else:
diff[i] = diff[i + 1]
min_diff = [(0) for i in range(a + b + c + 1)]
for i in range(a + b + c - 1, -1, -1):
min_diff[i] = min(min_diff[i + 1], diff[i])
p = float("inf")
for i in range(a + b + c + 1):
p = min(p, l1[i] + min_diff[i])
print(a + c + p) | ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | import sys
def main():
import sys
input = sys.stdin.readline
k1, k2, k3 = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))
dic = {}
for a in A:
dic[a] = 0
for b in B:
dic[b] = 1
for c in C:
dic[c] = 2
N = k1 + k2 + k3
dp = [([10**9] * 3) for _ in range(N + 1)]
dp[0][0] = 0
dp[0][1] = 0
dp[0][2] = 0
for i in range(1, N + 1):
dp[i][0] = min(dp[i][0], dp[i - 1][0] + (dic[i] != 0))
dp[i][1] = min(
dp[i][1], dp[i - 1][1] + (dic[i] != 1), dp[i - 1][0] + (dic[i] != 1)
)
dp[i][2] = min(
dp[i][2],
dp[i - 1][2] + (dic[i] != 2),
dp[i - 1][1] + (dic[i] != 2),
dp[i - 1][0] + (dic[i] != 2),
)
print(min(dp[-1]))
main() | IMPORT FUNC_DEF IMPORT ASSIGN VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP LIST BIN_OP NUMBER NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | import sys
input = sys.stdin.readline
K1, K2, K3 = map(int, input().split())
N = K1 + K2 + K3
A1 = list(map(int, input().split()))
A2 = list(map(int, input().split()))
A3 = list(map(int, input().split()))
A = [0] * N
for i, A_ in enumerate([A1, A2, A3], 1):
for a in A_:
A[a - 1] = i
a_12to3 = K1 + K2
ans = a_12to3
a_2to1 = 0
cnt_1 = 0
cnt_2 = 0
idx_12 = 0
for a in A:
if a == 3:
a_12to3 += 1
else:
a_12to3 -= 1
if a == 1:
cnt_1 += 1
if cnt_1 >= cnt_2:
a_2to1 += cnt_2
cnt_1 = cnt_2 = 0
else:
cnt_2 += 1
an = a_12to3 + a_2to1 + cnt_1
ans = min(ans, an)
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR FUNC_CALL VAR LIST VAR VAR VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | from sys import stdin
input = stdin.readline
r = list(map(int, input().split()))
n = sum(r)
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
mp = [0] * -~n
for i in a:
mp[i] = 0
for i in b:
mp[i] = 1
for i in c:
mp[i] = 2
inf = float("inf")
dp = [([inf] * -~n) for _ in range(3)]
for i in range(3):
dp[i][0] = 0
for j in range(1, n + 1):
dp[i][j] = dp[i][j - 1] + (mp[j] != i)
if i:
dp[i][j] = min(dp[i - 1][j], dp[i][j])
print(dp[2][n]) | ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP LIST VAR VAR VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER VAR VAR VAR IF VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | def LIS(X):
N = len(X)
P = [0] * N
M = [0] * (N + 1)
L = 0
for i in range(N):
lo = 1
hi = L
while lo <= hi:
mid = (lo + hi) // 2
if X[M[mid]] < X[i]:
lo = mid + 1
else:
hi = mid - 1
newL = lo
P[i] = M[newL - 1]
M[newL] = i
if newL > L:
L = newL
S = []
k = M[L]
for i in range(L - 1, -1, -1):
S.append(X[k])
k = P[k]
return len(S)
k1, k2, k3 = list(map(int, input().split()))
a = []
for _ in range(3):
a.extend(sorted(list(map(int, input().split()))))
print(k1 + k2 + k3 - LIS(a)) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | from sys import stdin, stdout
inf = 1000000000000007
def solve():
a1, b1, c1 = [int(i) for i in stdin.readline().split()]
k = a1 + b1 + c1
a = [(0) for i in range(k + 1)]
for i in stdin.readline().split():
a[int(i)] = 0
for i in stdin.readline().split():
a[int(i)] = 1
for i in stdin.readline().split():
a[int(i)] = 2
dp = [[inf for i in range(k + 1)] for j in range(3)]
dp[0][0], dp[1][0], dp[2][0] = 0, 0, 0
for i in range(1, k + 1):
for j in range(3):
if a[i] == j:
for l in range(0, j + 1):
dp[j][i] = min(dp[j][i], dp[l][i - 1])
else:
for l in range(0, j + 1):
dp[j][i] = min(dp[j][i], dp[l][i - 1] + 1)
print(min(dp[0][k], dp[1][k], dp[2][k]))
solve() | ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR EXPR FUNC_CALL VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | k1, k2, k3 = list(map(int, input().split(" ")))
n = k1 + k2 + k3
a = list(map(int, input().split(" ")))
b = list(map(int, input().split(" ")))
c = list(map(int, input().split(" ")))
B = list(range(k1 + k2 + k3))
for i in a:
B[i - 1] = 1
for i in b:
B[i - 1] = 2
for i in c:
B[i - 1] = 3
p1, p2, p3 = 0, 0, 0
inf = int(2 * 100000.0 + 3)
stateStorage = [[inf for i in range(3)] for j in range(2)]
for i in range(3):
stateStorage[n % 2][i] = 0
def fillMatrix(n):
for i in range(n - 1, -1, -1):
stateStorage[i % 2][0] = (
1
- int(B[i] == 1)
+ min(
stateStorage[(i + 1) % 2][0],
stateStorage[(i + 1) % 2][1],
stateStorage[(i + 1) % 2][2],
)
)
stateStorage[i % 2][1] = (
1
- int(B[i] == 2)
+ min(stateStorage[(i + 1) % 2][2], stateStorage[(i + 1) % 2][1])
)
stateStorage[i % 2][2] = 1 - int(B[i] == 3) + stateStorage[(i + 1) % 2][2]
return min(stateStorage[0][0], stateStorage[0][1], stateStorage[0][2])
ans = fillMatrix(n)
print(ans) | ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER RETURN FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | def input_sorted_int_list():
return sorted(map(int, input().split()))
def update_moves_count_list(moves_count_list, tasks_list, from_left_to_right):
p = 0
for i in range(n + 1):
if p < len(tasks_list) and i == tasks_list[p]:
p += 1
if from_left_to_right:
moves_count_list[i] += len(tasks_list) - p
else:
moves_count_list[i] += p
k1, k2, k3 = map(int, input().split())
n = k1 + k2 + k3
a = input_sorted_int_list()
b = input_sorted_int_list()
c = input_sorted_int_list()
left_moves_count = [0] * (n + 1)
right_moves_count = [0] * (n + 1)
update_moves_count_list(left_moves_count, a, True)
update_moves_count_list(left_moves_count, b, False)
update_moves_count_list(right_moves_count, b, True)
update_moves_count_list(right_moves_count, c, False)
r_cum_min = [None] * (n + 1)
r_cum_min[-1] = right_moves_count[n]
i = n - 1
while i >= 0:
r_cum_min[i] = min(right_moves_count[i], r_cum_min[i + 1])
i -= 1
answer = 1000000000.0
for i in range(n + 1):
if left_moves_count[i] + r_cum_min[i] < answer:
answer = left_moves_count[i] + r_cum_min[i]
print(answer) | FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NONE BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | import sys
input = sys.stdin.readline
k1, k2, k3 = map(int, input().split())
n = k1 + k2 + k3
a1 = list(map(int, input().split()))
a2 = list(map(int, input().split()))
a3 = list(map(int, input().split()))
b = [0] * n
for i in range(k2):
b[a2[i] - 1] = 1
for i in range(k3):
b[a3[i] - 1] = 2
cntr = [0] * 3
cntl = [0] * 3
for i in range(n):
cntr[b[i]] += 1
ans = k1 + k2
best_p = 0
for i in range(n):
cntl[b[i]] += 1
cntr[b[i]] -= 1
best_p = max(best_p, cntl[0] - cntl[1])
ans = min(ans, cntl[0] + cntl[2] + cntr[0] + cntr[1] - best_p)
print(ans) | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | import sys
def main():
n1, n2, n3 = readIntArr()
n = n1 + n2 + n3
a = readIntArr()
b = readIntArr()
c = readIntArr()
origPos = [(-1) for _ in range(n + 1)]
for x in a:
origPos[x] = 1
for x in b:
origPos[x] = 2
for x in c:
origPos[x] = 3
dp = makeArr(inf, [n + 1, 4])
dp[0][1] = 0
dp[0][2] = 0
dp[0][3] = 0
for i in range(1, n + 1):
for j in range(1, 4):
dp[i][j] = min(dp[i][j], dp[i][j - 1])
if origPos[i] != j:
extraMove = 1
else:
extraMove = 0
dp[i][j] = min(dp[i][j], dp[i - 1][j] + extraMove)
ans = dp[n][3]
print(ans)
return
input = lambda: sys.stdin.readline().rstrip("\r\n")
def oneLineArrayPrint(arr):
print(" ".join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print("\n".join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print("\n".join([" ".join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
def makeArr(defaultVal, dimensionArr):
dv = defaultVal
da = dimensionArr
if len(da) == 1:
return [dv for _ in range(da[0])]
else:
return [makeArr(dv, da[1:]) for _ in range(da[0])]
def queryInteractive(x, y):
print("? {} {}".format(x, y))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print("! {}".format(ans))
sys.stdout.flush()
inf = float("inf")
MOD = 10**9 + 7
for _abc in range(1):
main() | IMPORT FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR LIST BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN VAR VAR FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR EXPR FUNC_CALL VAR RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | k = [int(i) for i in input().split(" ")]
n = sum(k)
a = [0] * n
for I in range(3):
line = [int(i) for i in input().split(" ")]
for item in line:
a[item - 1] = I
dp = [1] * n
cur = [-1, -1, -1]
for i in range(n):
for j in range(a[i] + 1):
if cur[j] >= 0:
dp[i] = max(dp[i], 1 + dp[cur[j]])
cur[a[i]] = i
print(n - max(dp)) | ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING FOR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER VAR VAR VAR ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR |
A team of three programmers is going to play a contest. The contest consists of $n$ problems, numbered from $1$ to $n$. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$. The second one has received problems $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$. The third one has received all remaining problems ($a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
-----Input-----
The first line contains three integers $k_1, k_2$ and $k_3$ ($1 \le k_1, k_2, k_3 \le 2 \cdot 10^5, k_1 + k_2 + k_3 \le 2 \cdot 10^5$) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains $k_1$ integers $a_{1, 1}, a_{1, 2}, \dots, a_{1, k_1}$ — the problems initially taken by the first participant.
The third line contains $k_2$ integers $a_{2, 1}, a_{2, 2}, \dots, a_{2, k_2}$ — the problems initially taken by the second participant.
The fourth line contains $k_3$ integers $a_{3, 1}, a_{3, 2}, \dots, a_{3, k_3}$ — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer $a_{i, j}$ meets the condition $1 \le a_{i, j} \le n$, where $n = k_1 + k_2 + k_3$.
-----Output-----
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
-----Examples-----
Input
2 1 2
3 1
4
2 5
Output
1
Input
3 2 1
3 2 1
5 4
6
Output
0
Input
2 1 3
5 6
4
1 2 3
Output
3
Input
1 5 1
6
5 1 2 4 7
3
Output
2
-----Note-----
In the first example the third contestant should give the problem $2$ to the first contestant, so the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $1$ remaining problem.
In the second example the distribution of problems is already valid: the first contestant has $3$ first problems, the third contestant has $1$ last problem, and the second contestant has $2$ remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant. | import sys
readline = sys.stdin.readline
readlines = sys.stdin.readlines
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep="\n")
def solve():
n1, n2, n3 = nm()
n = n1 + n2 + n3
g1 = nl()
g2 = nl()
g3 = nl()
f1 = [0] * (n + 1)
f2 = [0] * (n + 1)
f3 = [0] * (n + 1)
for x in g1:
f1[x] = 1
for x in g2:
f2[x] = 1
for x in g3:
f3[x] = 1
for i in range(n):
f1[i + 1] += f1[i]
f2[i + 1] += f2[i]
f3[i + 1] += f3[i]
h12 = [(f2[i] - f1[i]) for i in range(n + 1)]
h23 = [(f3[i] - f2[i]) for i in range(n + 1)]
for i in range(n - 1, -1, -1):
h23[i] = min(h23[i], h23[i + 1])
ans = n1 + n2 + min(h12[i] + h23[i] for i in range(n + 1))
print(ans)
return
solve() | IMPORT ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR STRING FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR |
A sequence of round and square brackets is given. You can change the sequence by performing the following operations:
change the direction of a bracket from opening to closing and vice versa without changing the form of the bracket: i.e. you can change '(' to ')' and ')' to '('; you can change '[' to ']' and ']' to '['. The operation costs $0$ burles.
change any square bracket to round bracket having the same direction: i.e. you can change '[' to '(' but not from '(' to '['; similarly, you can change ']' to ')' but not from ')' to ']'. The operation costs $1$ burle.
The operations can be performed in any order any number of times.
You are given a string $s$ of the length $n$ and $q$ queries of the type "l r" where $1 \le l < r \le n$. For every substring $s[l \dots r]$, find the minimum cost to pay to make it a correct bracket sequence. It is guaranteed that the substring $s[l \dots r]$ has an even length.
The queries must be processed independently, i.e. the changes made in the string for the answer to a question $i$ don't affect the queries $j$ ($j > i$). In other words, for every query, the substring $s[l \dots r]$ is given from the initially given string $s$.
A correct bracket sequence is a sequence that can be built according the following rules:
an empty sequence is a correct bracket sequence;
if "s" is a correct bracket sequence, the sequences "(s)" and "[s]" are correct bracket sequences.
if "s" and "t" are correct bracket sequences, the sequence "st" (the concatenation of the sequences) is a correct bracket sequence.
E.g. the sequences "", "(()[])", "[()()]()" and "(())()" are correct bracket sequences whereas "(", "[(])" and ")))" are not.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 100$) — the number of test cases. Then $t$ test cases follow.
For each test case, the first line contains a non-empty string $s$ containing only round ('(', ')') and square ('[', ']') brackets. The length of the string doesn't exceed $10^6$. The string contains at least $2$ characters.
The second line contains one integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.
Then $q$ lines follow, each of them contains two integers $l$ and $r$ ($1 \le l < r \le n$ where $n$ is the length of $s$). It is guaranteed that the substring $s[l \dots r]$ has even length.
It is guaranteed that the sum of the lengths of all strings given in all test cases doesn't exceed $10^6$. The sum of all $q$ given in all test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case output in a separate line for each query one integer $x$ ($x \ge 0$) — the minimum cost to pay to make the given substring a correct bracket sequence.
-----Examples-----
Input
3
([))[)()][]]
3
1 12
4 9
3 6
))))))
2
2 3
1 4
[]
1
1 2
Output
0
2
1
0
0
0
-----Note-----
Consider the first test case. The first query describes the whole given string, the string can be turned into the following correct bracket sequence: "([()])()[[]]". The forms of the brackets aren't changed so the cost of changing is $0$.
The second query describes the substring ")[)()]". It may be turned into "(()())", the cost is equal to $2$.
The third query describes the substring "))[)". It may be turned into "()()", the cost is equal to $1$.
The substrings of the second test case contain only round brackets. It's possible to prove that any sequence of round brackets having an even length may be turned into a correct bracket sequence for the cost of $0$ burles.
In the third test case, the single query describes the string "[]" that is already a correct bracket sequence. | import sys
input = sys.stdin.readline
t = int(input())
for tests in range(t):
X = input().strip()
n = len(X)
R = [0]
L = [0]
for i in range(n):
if i % 2 == 0:
if X[i] == "[" or X[i] == "]":
R.append(R[-1] + 1)
else:
R.append(R[-1])
L.append(L[-1])
else:
if X[i] == "[" or X[i] == "]":
L.append(L[-1] + 1)
else:
L.append(L[-1])
R.append(R[-1])
q = int(input())
for queries in range(q):
l, r = map(int, input().split())
RS = R[r] - R[l - 1]
LS = L[r] - L[l - 1]
print(abs(RS - LS)) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER IF VAR VAR STRING VAR VAR STRING EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR |
A sequence of round and square brackets is given. You can change the sequence by performing the following operations:
change the direction of a bracket from opening to closing and vice versa without changing the form of the bracket: i.e. you can change '(' to ')' and ')' to '('; you can change '[' to ']' and ']' to '['. The operation costs $0$ burles.
change any square bracket to round bracket having the same direction: i.e. you can change '[' to '(' but not from '(' to '['; similarly, you can change ']' to ')' but not from ')' to ']'. The operation costs $1$ burle.
The operations can be performed in any order any number of times.
You are given a string $s$ of the length $n$ and $q$ queries of the type "l r" where $1 \le l < r \le n$. For every substring $s[l \dots r]$, find the minimum cost to pay to make it a correct bracket sequence. It is guaranteed that the substring $s[l \dots r]$ has an even length.
The queries must be processed independently, i.e. the changes made in the string for the answer to a question $i$ don't affect the queries $j$ ($j > i$). In other words, for every query, the substring $s[l \dots r]$ is given from the initially given string $s$.
A correct bracket sequence is a sequence that can be built according the following rules:
an empty sequence is a correct bracket sequence;
if "s" is a correct bracket sequence, the sequences "(s)" and "[s]" are correct bracket sequences.
if "s" and "t" are correct bracket sequences, the sequence "st" (the concatenation of the sequences) is a correct bracket sequence.
E.g. the sequences "", "(()[])", "[()()]()" and "(())()" are correct bracket sequences whereas "(", "[(])" and ")))" are not.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 100$) — the number of test cases. Then $t$ test cases follow.
For each test case, the first line contains a non-empty string $s$ containing only round ('(', ')') and square ('[', ']') brackets. The length of the string doesn't exceed $10^6$. The string contains at least $2$ characters.
The second line contains one integer $q$ ($1 \le q \le 2 \cdot 10^5$) — the number of queries.
Then $q$ lines follow, each of them contains two integers $l$ and $r$ ($1 \le l < r \le n$ where $n$ is the length of $s$). It is guaranteed that the substring $s[l \dots r]$ has even length.
It is guaranteed that the sum of the lengths of all strings given in all test cases doesn't exceed $10^6$. The sum of all $q$ given in all test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case output in a separate line for each query one integer $x$ ($x \ge 0$) — the minimum cost to pay to make the given substring a correct bracket sequence.
-----Examples-----
Input
3
([))[)()][]]
3
1 12
4 9
3 6
))))))
2
2 3
1 4
[]
1
1 2
Output
0
2
1
0
0
0
-----Note-----
Consider the first test case. The first query describes the whole given string, the string can be turned into the following correct bracket sequence: "([()])()[[]]". The forms of the brackets aren't changed so the cost of changing is $0$.
The second query describes the substring ")[)()]". It may be turned into "(()())", the cost is equal to $2$.
The third query describes the substring "))[)". It may be turned into "()()", the cost is equal to $1$.
The substrings of the second test case contain only round brackets. It's possible to prove that any sequence of round brackets having an even length may be turned into a correct bracket sequence for the cost of $0$ burles.
In the third test case, the single query describes the string "[]" that is already a correct bracket sequence. | import sys
input = sys.stdin.readline
for _ in range(int(input())):
s = input().strip()
n = len(s)
odd = [0]
even = [0]
for i in range(n):
if s[i] == "[" or s[i] == "]":
if i % 2 == 0:
even.append(even[-1] + 1)
odd.append(odd[-1])
else:
odd.append(odd[-1] + 1)
even.append(even[-1])
else:
even.append(even[-1])
odd.append(odd[-1])
for query in range(int(input())):
l, r = [int(x) for x in input().split(" ")]
oddS = odd[r] - odd[l - 1]
evenS = even[r] - even[l - 1]
print(abs(oddS - evenS)) | IMPORT ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR STRING IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR |
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
You are given a tree with $N$ nodes (numbered $1$ through $N$), rooted at node $1$. For each valid $i$, node $i$ has a value $a_{i}$ written on it.
An undirected simple path between any two nodes $u$ and $v$ is said to be vertical if $u=v$ or $u$ is an ancestor of $v$ or $v$ is an ancestor of $u$. Let's define a *vertical partition* of the tree as a set of vertical paths such that each node belongs to exactly one of these paths.
You are also given a sequence of $N$ integers $b_{1}, b_{2}, \ldots, b_{N}$. A vertical partition is *good* if, for each of its paths, we can permute the values written on the nodes in this path, and this can be done in such a way that we reach a state where for each valid $i$, the value written on node $i$ is $b_{i}$.
The difficulty of your task is described by a parameter $S$. If $S=1$, your task is only to determine whether at least one good vertical partition exists. If $S=2$, you are required to find the number of good vertical partitions modulo $1,000,000,007$ ($10^{9}+7$).
------ Input ------
The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
The first line of each test case contains two space-separated integers $N$ and $S$.
Each of the next $N-1$ lines contains two space-separated integers $u$ and $v$ denoting that nodes $u$ and $v$ are connected by an edge.
The next line contains $N$ space-separated integers $a_{1}, a_{2}, \ldots, a_{N}$.
The line after that contains $N$ space-separated integers $b_{1}, b_{2}, \ldots, b_{N}$.
------ Output ------
For each test case, print a single line containing one integer:
If $S=1$, this integer should be $1$ if a good vertical partition exists or $0$ if it does not exist.
If $S=2$, this integer should be the number of good vertical partitions modulo $1,000,000,007$ ($10^{9}+7$).
------ Constraints ------
$1 ≤ T ≤ 10^{6}$
$1 ≤ N ≤ 10^{5}$
$S \in \{1, 2\}$
$1 ≤ u, v ≤ N$
the graph described on the input is a tree
$1 ≤ a_{i}, b_{i} ≤ 10^{6}$ for each valid $i$
the sum of $N$ over all test cases does not exceed $10^{6}$
------ Subtasks ------
Subtask #1 (40 points, time limit 1 seconds): the sum of $N$ over all test cases does not exceed $1,000$
Subtask #2 (30 points, time limit 2 seconds): $S = 1$
Subtask #3 (30 points, time limit 2 seconds): original constraints
----- Sample Input 1 ------
4
3 2
1 2
2 3
4 5 6
4 6 5
6 2
1 2
1 3
2 4
3 5
3 6
10 20 30 40 50 60
10 40 50 20 30 60
6 1
1 2
1 3
2 4
3 5
3 6
10 20 30 40 50 60
10 40 50 20 30 60
1 2
1
2
----- Sample Output 1 ------
2
3
1
0
----- explanation 1 ------
Example case 1: The good vertical partitions are $\{[1], [2, 3]\}$ and $\{[1, 2, 3]\}$.
Example case 2: The good vertical partitions are:
- $\{[1, 2, 4], [3, 5], [6]\}$
- $\{[1], [2, 4], [3, 5], [6]\}$
- $\{[1, 3, 5], [2, 4], [6]\}$
Example case 3: The same as example case 2, but with $S=1$.
Example case 4: There is no good vertical partition. | import sys
sys.setrecursionlimit(100100)
def solve():
n, s = map(int, input().split())
children = []
for i in range(n):
children.append([])
for i in range(n - 1):
u, v = map(int, input().split())
children[u - 1].append(v - 1)
children[v - 1].append(u - 1)
a = list(map(int, input().split()))
b = list(map(int, input().split()))
vis = [False] * n
req, avail, comb = find_path(0, children, a, b, vis)
if req or req is None:
print(0)
elif s == 1:
print(1)
else:
print(comb % MODULO)
MODULO = 1000000007
def find_path(node, children, a, b, vis):
connected = False
vis[node] = True
ret_req, ret_avail, children_comb = None, None, 1
for child in children[node]:
if vis[child]:
continue
req, avail, comb = find_path(child, children, a, b, vis)
if req:
if connected:
return None, None, None
connected = True
if a[node] != b[node]:
if b[node] in avail:
if avail[b[node]] == 1:
del avail[b[node]]
else:
avail[b[node]] -= 1
elif b[node] in req:
req[b[node]] += 1
else:
req[b[node]] = 1
if a[node] in req:
if req[a[node]] == 1:
del req[a[node]]
else:
req[a[node]] -= 1
elif a[node] in avail:
avail[a[node]] += 1
else:
avail[a[node]] = 1
ret_req, ret_avail = req, avail
elif req is None:
return None, None, None
children_comb = children_comb * comb
if not connected:
ret_req, ret_avail = {}, {}
if a[node] != b[node]:
ret_req[b[node]] = 1
ret_avail[a[node]] = 1
if node != 0:
children_comb = children_comb * len(children[node])
else:
children_comb = children_comb * (len(children[node]) + 1)
return ret_req, ret_avail, children_comb
t = int(input())
for i in range(t):
solve() | IMPORT EXPR FUNC_CALL VAR NUMBER FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR VAR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR IF VAR VAR NONE EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR NONE NONE NUMBER FOR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR IF VAR IF VAR RETURN NONE NONE NONE ASSIGN VAR NUMBER IF VAR VAR VAR VAR IF VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NONE RETURN NONE NONE NONE ASSIGN VAR BIN_OP VAR VAR IF VAR ASSIGN VAR VAR DICT DICT IF VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER RETURN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
You are given a tree with $N$ nodes (numbered $1$ through $N$), rooted at node $1$. For each valid $i$, node $i$ has a value $a_{i}$ written on it.
An undirected simple path between any two nodes $u$ and $v$ is said to be vertical if $u=v$ or $u$ is an ancestor of $v$ or $v$ is an ancestor of $u$. Let's define a *vertical partition* of the tree as a set of vertical paths such that each node belongs to exactly one of these paths.
You are also given a sequence of $N$ integers $b_{1}, b_{2}, \ldots, b_{N}$. A vertical partition is *good* if, for each of its paths, we can permute the values written on the nodes in this path, and this can be done in such a way that we reach a state where for each valid $i$, the value written on node $i$ is $b_{i}$.
The difficulty of your task is described by a parameter $S$. If $S=1$, your task is only to determine whether at least one good vertical partition exists. If $S=2$, you are required to find the number of good vertical partitions modulo $1,000,000,007$ ($10^{9}+7$).
------ Input ------
The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
The first line of each test case contains two space-separated integers $N$ and $S$.
Each of the next $N-1$ lines contains two space-separated integers $u$ and $v$ denoting that nodes $u$ and $v$ are connected by an edge.
The next line contains $N$ space-separated integers $a_{1}, a_{2}, \ldots, a_{N}$.
The line after that contains $N$ space-separated integers $b_{1}, b_{2}, \ldots, b_{N}$.
------ Output ------
For each test case, print a single line containing one integer:
If $S=1$, this integer should be $1$ if a good vertical partition exists or $0$ if it does not exist.
If $S=2$, this integer should be the number of good vertical partitions modulo $1,000,000,007$ ($10^{9}+7$).
------ Constraints ------
$1 ≤ T ≤ 10^{6}$
$1 ≤ N ≤ 10^{5}$
$S \in \{1, 2\}$
$1 ≤ u, v ≤ N$
the graph described on the input is a tree
$1 ≤ a_{i}, b_{i} ≤ 10^{6}$ for each valid $i$
the sum of $N$ over all test cases does not exceed $10^{6}$
------ Subtasks ------
Subtask #1 (40 points, time limit 1 seconds): the sum of $N$ over all test cases does not exceed $1,000$
Subtask #2 (30 points, time limit 2 seconds): $S = 1$
Subtask #3 (30 points, time limit 2 seconds): original constraints
----- Sample Input 1 ------
4
3 2
1 2
2 3
4 5 6
4 6 5
6 2
1 2
1 3
2 4
3 5
3 6
10 20 30 40 50 60
10 40 50 20 30 60
6 1
1 2
1 3
2 4
3 5
3 6
10 20 30 40 50 60
10 40 50 20 30 60
1 2
1
2
----- Sample Output 1 ------
2
3
1
0
----- explanation 1 ------
Example case 1: The good vertical partitions are $\{[1], [2, 3]\}$ and $\{[1, 2, 3]\}$.
Example case 2: The good vertical partitions are:
- $\{[1, 2, 4], [3, 5], [6]\}$
- $\{[1], [2, 4], [3, 5], [6]\}$
- $\{[1, 3, 5], [2, 4], [6]\}$
Example case 3: The same as example case 2, but with $S=1$.
Example case 4: There is no good vertical partition. | import sys
sys.setrecursionlimit(10**6)
N = 10**5 + 10
global n, s
mod = 10**9 + 7
global ok
graph = [[] for i in range(N)]
spec = [(0) for i in range(N)]
a, b = [(0) for i in range(N)], [(0) for i in range(N)]
dp = [(0) for i in range(N)]
bal = [{} for i in range(N)]
dummy = {}
def dfs1(node, par, cntr, freq):
global ok
cnt = 0
for to in graph[node]:
if to == par:
continue
tempo, t_freq = dfs1(to, node, bal[to], 0)
if not ok:
return dummy, -1
if t_freq == 0:
spec[to] = 1
tempo.clear()
else:
cnt += 1
cntr = tempo
freq = t_freq
if cnt > 1:
ok = False
return dummy, -1
aa = a[node]
bb = b[node]
if cntr.get(aa, 0) == 0:
freq += 1
cntr[aa] = cntr.get(aa, 0) - 1
else:
cntr[aa] = cntr.get(aa, 0) - 1
if cntr.get(aa, 0) == 0:
freq -= 1
if cntr.get(bb, 0) == 0:
freq += 1
cntr[bb] = cntr.get(bb, 0) + 1
else:
cntr[bb] = cntr.get(bb, 0) + 1
if cntr.get(bb, 0) == 0:
freq -= 1
return cntr, freq
def dfs2(node, par):
global ok
dp[node] = 1
cnt = 0
temp = True
for to in graph[node]:
if to == par:
continue
if not spec[to]:
temp = False
dfs2(to, node)
dp[node] *= dp[to]
dp[node] %= mod
cnt += 1
if temp:
dp[node] *= cnt + 1
dp[node] %= mod
for _ in range(int(input())):
global ok
n, s = [int(i) for i in input().split()]
ok = True
for i in range(n - 1):
u, v = [int(i) for i in input().split()]
u -= 1
v -= 1
graph[u].append(v)
graph[v].append(u)
t1 = [int(i) for i in input().split()]
t2 = [int(i) for i in input().split()]
for i in range(n):
a[i] = t1[i]
b[i] = t2[i]
not_req, freq = dfs1(0, -1, bal[0], 0)
if freq != 0:
ok = False
else:
spec[0] = True
if not ok:
print(0)
elif s == 1:
print(1)
else:
dfs2(0, -1)
print(dp[0])
for i in range(n):
graph[i].clear()
dp[i] = 0
spec[i] = 0
bal[i].clear() | IMPORT EXPR FUNC_CALL VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR LIST VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR DICT VAR FUNC_CALL VAR VAR ASSIGN VAR DICT FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER IF VAR RETURN VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR VAR IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER NUMBER VAR NUMBER NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER NUMBER IF VAR EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Read problem statements in [Bengali], [Mandarin Chinese], [Russian], and [Vietnamese] as well.
You are given a tree with $N$ nodes (numbered $1$ through $N$), rooted at node $1$. For each valid $i$, node $i$ has a value $a_{i}$ written on it.
An undirected simple path between any two nodes $u$ and $v$ is said to be vertical if $u=v$ or $u$ is an ancestor of $v$ or $v$ is an ancestor of $u$. Let's define a *vertical partition* of the tree as a set of vertical paths such that each node belongs to exactly one of these paths.
You are also given a sequence of $N$ integers $b_{1}, b_{2}, \ldots, b_{N}$. A vertical partition is *good* if, for each of its paths, we can permute the values written on the nodes in this path, and this can be done in such a way that we reach a state where for each valid $i$, the value written on node $i$ is $b_{i}$.
The difficulty of your task is described by a parameter $S$. If $S=1$, your task is only to determine whether at least one good vertical partition exists. If $S=2$, you are required to find the number of good vertical partitions modulo $1,000,000,007$ ($10^{9}+7$).
------ Input ------
The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
The first line of each test case contains two space-separated integers $N$ and $S$.
Each of the next $N-1$ lines contains two space-separated integers $u$ and $v$ denoting that nodes $u$ and $v$ are connected by an edge.
The next line contains $N$ space-separated integers $a_{1}, a_{2}, \ldots, a_{N}$.
The line after that contains $N$ space-separated integers $b_{1}, b_{2}, \ldots, b_{N}$.
------ Output ------
For each test case, print a single line containing one integer:
If $S=1$, this integer should be $1$ if a good vertical partition exists or $0$ if it does not exist.
If $S=2$, this integer should be the number of good vertical partitions modulo $1,000,000,007$ ($10^{9}+7$).
------ Constraints ------
$1 ≤ T ≤ 10^{6}$
$1 ≤ N ≤ 10^{5}$
$S \in \{1, 2\}$
$1 ≤ u, v ≤ N$
the graph described on the input is a tree
$1 ≤ a_{i}, b_{i} ≤ 10^{6}$ for each valid $i$
the sum of $N$ over all test cases does not exceed $10^{6}$
------ Subtasks ------
Subtask #1 (40 points, time limit 1 seconds): the sum of $N$ over all test cases does not exceed $1,000$
Subtask #2 (30 points, time limit 2 seconds): $S = 1$
Subtask #3 (30 points, time limit 2 seconds): original constraints
----- Sample Input 1 ------
4
3 2
1 2
2 3
4 5 6
4 6 5
6 2
1 2
1 3
2 4
3 5
3 6
10 20 30 40 50 60
10 40 50 20 30 60
6 1
1 2
1 3
2 4
3 5
3 6
10 20 30 40 50 60
10 40 50 20 30 60
1 2
1
2
----- Sample Output 1 ------
2
3
1
0
----- explanation 1 ------
Example case 1: The good vertical partitions are $\{[1], [2, 3]\}$ and $\{[1, 2, 3]\}$.
Example case 2: The good vertical partitions are:
- $\{[1, 2, 4], [3, 5], [6]\}$
- $\{[1], [2, 4], [3, 5], [6]\}$
- $\{[1, 3, 5], [2, 4], [6]\}$
Example case 3: The same as example case 2, but with $S=1$.
Example case 4: There is no good vertical partition. | MOD = int(1000000000.0 + 7)
class TreeNode:
def __init__(self, val=0, children=[], parent=None):
self.v = val
self.c = children
self.p = parent
self.a = 0
self.b = 0
self.count = 1
def __repr__(self):
return self.__str__()
def __str__(self):
return str(self.v)
def solve():
n, ss = map(int, input().split())
nodes = [None]
for i in range(1, n + 1):
nodes.append(TreeNode(i, []))
edges = {}
for i in range(1, n):
s = input()
u, v = map(int, s.split())
if u in edges:
edges[u].append(v)
else:
edges[u] = [v]
if v in edges:
edges[v].append(u)
else:
edges[v] = [u]
st = [nodes[1]]
done = {}
if len(edges) > 0:
while len(st) > 0:
node = st.pop()
done[node.v] = True
for c in edges[node.v]:
if not c in done:
node.c.append(nodes[c])
nodes[c].p = node
st.append(nodes[c])
a = list(map(int, input().split()))
b = list(map(int, input().split()))
for i in range(1, n + 1):
nodes[i].a = a[i - 1]
nodes[i].b = b[i - 1]
levels = [[nodes[1]]]
while len(levels[-1]) > 0:
l = []
for node in levels[-1]:
l.extend(node.c)
levels.append(l)
levels.pop()
ans = 1
passed = {}
for i in range(len(levels) - 1, -1, -1):
for node in levels[i]:
if node.v in passed:
continue
ans = ans * (len(node.c) + 1) % MOD
if node.a == node.b:
continue
cur = node
diff = {}
while True:
if cur.v in passed:
print(0)
return
passed[cur.v] = True
if cur.a in diff:
diff[cur.a] -= 1
if diff[cur.a] == 0:
del diff[cur.a]
else:
diff[cur.a] = -1
if cur.b in diff:
diff[cur.b] += 1
if diff[cur.b] == 0:
del diff[cur.b]
else:
diff[cur.b] = 1
temp_v = cur.v
if cur.v == 1 and len(diff) > 0:
print(0)
return
cur = cur.p
if len(diff) == 0:
break
if ss == 2:
print(ans)
elif ans >= 1:
print(1)
else:
print(0)
t = int(input())
for _ in range(t):
solve() | ASSIGN VAR FUNC_CALL VAR BIN_OP NUMBER NUMBER CLASS_DEF FUNC_DEF NUMBER LIST NONE ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NONE FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR LIST ASSIGN VAR DICT FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR LIST VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR LIST VAR ASSIGN VAR LIST VAR NUMBER ASSIGN VAR DICT IF FUNC_CALL VAR VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR LIST LIST VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR LIST FOR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR DICT FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER FOR VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR DICT WHILE NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
step = 0
l, r = -1, 0
for i, j in sorted(clips):
if r >= T:
return step
if i > r:
return -1
if l < i <= r:
step += 1
l = r
r = max(r, j)
return step if r >= T else -1 | CLASS_DEF FUNC_DEF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR RETURN VAR IF VAR VAR RETURN NUMBER IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR VAR NUMBER VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
clips.sort()
n = len(clips)
dp = [float("inf")] * (T + 1)
dp[0] = 0
for i in range(1, T + 1):
for j in range(n):
if clips[j][0] <= i <= clips[j][1]:
dp[i] = min(dp[i], 1 + dp[clips[j][0]])
if dp[T] == float("inf"):
return -1
return dp[-1] | CLASS_DEF FUNC_DEF VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST FUNC_CALL VAR STRING BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER VAR VAR VAR NUMBER IF VAR VAR FUNC_CALL VAR STRING RETURN NUMBER RETURN VAR NUMBER VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
clips.sort(key=lambda x: [x[0], -x[1]])
res = 0
r = 0
new_r = 0
for s, e in clips:
if s <= r:
new_r = max(new_r, e)
elif s > new_r:
return -1
else:
res += 1
r = new_r
new_r = max(new_r, e)
if new_r >= T:
break
if new_r < T:
return -1
if r < T and new_r >= T:
res += 1
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR EXPR FUNC_CALL VAR LIST VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR RETURN NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR IF VAR VAR RETURN NUMBER IF VAR VAR VAR VAR VAR NUMBER RETURN VAR VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
clips.sort()
i = 0
end = 0
res = 0
while end < T:
current_end = 0
while i < len(clips) and clips[i][0] <= end:
current_end = max(current_end, clips[i][1])
i += 1
if current_end <= end:
return -1
end = current_end
res += 1
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR RETURN NUMBER ASSIGN VAR VAR VAR NUMBER RETURN VAR VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
table = [101] * 101
table[0] = 0
changed = True
while changed:
changed = False
for i, j in clips:
for k in range(i, j + 1):
if table[i] + 1 < table[k]:
table[k] = table[i] + 1
changed = True
if table[T] < 101:
return table[T]
else:
return -1 | CLASS_DEF FUNC_DEF VAR VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR NUMBER FOR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR NUMBER RETURN VAR VAR RETURN NUMBER VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips, T):
end, end2, res = -1, 0, 0
for i, j in sorted(clips):
if end2 >= T or i > end2:
break
elif end < i <= end2:
res, end = res + 1, end2
end2 = max(end2, j)
return res if end2 >= T else -1 | CLASS_DEF FUNC_DEF ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR IF VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR VAR NUMBER |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
clips.sort(key=lambda x: x[0])
n = len(clips)
memo = {}
def recur(i, interval):
if (i, interval) in memo:
return memo[i, interval]
if interval >= T:
return 0
if i == n:
return sys.maxsize
if interval >= clips[i][0] and interval < clips[i][1]:
ans = min(recur(i + 1, clips[i][1]) + 1, recur(i + 1, interval))
else:
ans = recur(i + 1, interval)
memo[i, interval] = ans
return ans
fnl = recur(0, 0)
return -1 if fnl == sys.maxsize else fnl | CLASS_DEF FUNC_DEF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR DICT FUNC_DEF IF VAR VAR VAR RETURN VAR VAR VAR IF VAR VAR RETURN NUMBER IF VAR VAR RETURN VAR IF VAR VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR NUMBER NUMBER RETURN VAR VAR NUMBER VAR VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
dp = [T + 1] * (T + 1)
dp[0] = 0
for i in range(1, T + 1):
if dp[i - 1] >= T:
break
for start, end in clips:
if start <= i <= end:
dp[i] = min(dp[i], dp[start] + 1)
print(dp)
if dp[T] == T + 1:
return -1
else:
return dp[T] | CLASS_DEF FUNC_DEF VAR VAR VAR VAR ASSIGN VAR BIN_OP LIST BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR FOR VAR VAR VAR IF VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR NUMBER RETURN NUMBER RETURN VAR VAR VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
clips.sort(key=lambda x: x[0])
dp = [[0] * (len(clips) + 1)] + [
([float("inf")] * (len(clips) + 1)) for _ in range(T)
]
for row in range(1, len(dp)):
for col in range(1, len(dp[0])):
dp[row][col] = dp[row][col - 1]
if clips[col - 1][0] <= row - 1 < clips[col - 1][1]:
dp[row][col] = min(dp[row][col], 1 + dp[clips[col - 1][0]][col])
ans = dp[-1][-1] if dp[-1][-1] != float("inf") else -1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP LIST FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR VAR NUMBER NUMBER FUNC_CALL VAR STRING VAR NUMBER NUMBER NUMBER RETURN VAR VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
last_end = 0
clip_count = 0
while last_end < T:
if clip_count >= len(clips):
return -1
max_end = -1
for clip in clips:
if clip[0] <= last_end:
if clip[1] > max_end:
max_end = clip[1]
if max_end >= 0:
last_end = max_end
clip_count += 1
else:
return -1
return clip_count | CLASS_DEF FUNC_DEF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR IF VAR NUMBER VAR ASSIGN VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR NUMBER RETURN NUMBER RETURN VAR VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
dp = [float(inf)] * 101
clips.sort()
if clips[0][0] > 0:
return -1
dp[0] = 0
for i in range(len(clips)):
clip = clips[i]
if clip[0] != 0 and dp[clip[0]] == float(inf):
break
min_so_far = dp[clip[0]]
for k in range(clip[0], clip[1] + 1):
dp[k] = min(dp[k], min_so_far + 1)
if dp[T] == inf:
return -1
return dp[T] | CLASS_DEF FUNC_DEF VAR VAR VAR VAR ASSIGN VAR BIN_OP LIST FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR IF VAR NUMBER NUMBER NUMBER RETURN NUMBER ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER NUMBER VAR VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR VAR RETURN NUMBER RETURN VAR VAR VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
dp = [T + 1] * (T + 1)
dp[0] = 0
i = 1
while i <= T and dp[i - 1] < T:
for c in clips:
if c[0] <= i and i <= c[1]:
dp[i] = min(dp[i], dp[c[0]] + 1)
i += 1
return dp[T] if dp[T] != T + 1 else -1 | CLASS_DEF FUNC_DEF VAR VAR VAR VAR ASSIGN VAR BIN_OP LIST BIN_OP VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR BIN_OP VAR NUMBER VAR FOR VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER NUMBER VAR NUMBER RETURN VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR |
You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. We can cut these clips into segments freely: for example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7].
Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 10
Output: 3
Explanation:
We take the clips [0,2], [8,10], [1,9]; a total of 3 clips.
Then, we can reconstruct the sporting event as follows:
We cut [1,9] into segments [1,2] + [2,8] + [8,9].
Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10].
Example 2:
Input: clips = [[0,1],[1,2]], T = 5
Output: -1
Explanation:
We can't cover [0,5] with only [0,1] and [1,2].
Example 3:
Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100 | class Solution:
def videoStitching(self, clips: List[List[int]], T: int) -> int:
end, end2, cnt = -1, 0, 0
for s, e in sorted(clips):
if end2 >= T or s > end2:
break
elif end < s <= end2:
cnt += 1
end = end2
end2 = max(end2, e)
return cnt if end2 >= T else -1 | CLASS_DEF FUNC_DEF VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR VAR NUMBER VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
s, t = 0, 0
for x in a:
t += x
if t < 0:
t = 0
s = max(s, t)
pr, t = 0, 0
for x in a:
t += x
pr = max(pr, t)
su, t = 0, 0
for x in reversed(a):
t += x
su = max(su, t)
sb = sum([x for x in b if x > 0])
print(max(s, pr + sb, su + sb)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | def solve(N, A, M, B):
cum_sum = 0
start_sum = 0
end_sum = 0
for i in A:
if cum_sum >= 0:
cum_sum += i
start_sum = max(cum_sum, start_sum)
cum_sum = 0
for i in A[::-1]:
if cum_sum >= 0:
cum_sum += i
end_sum = max(cum_sum, end_sum)
sum_to_be_added = 0
cum_sum = 0
mid_sum = 0
for i in A[1:-1]:
if cum_sum >= 0:
cum_sum += i
end_sum = max(cum_sum, end_sum)
for i in B:
if i > 0:
sum_to_be_added += i
return sum_to_be_added + max(end_sum, mid_sum, start_sum)
return max(start_sum, end_sum)
T = int(input())
while T:
N = int(input())
A = [int(x) for x in input().split()]
M = int(input())
B = [int(x) for x in input().split()]
ans = solve(N, A, M, B)
print(ans)
T -= 1 | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR NUMBER NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR VAR IF VAR NUMBER VAR VAR RETURN BIN_OP VAR FUNC_CALL VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | def kadane(arr):
res = cur = 0
for i in arr:
cur += i
res = max(res, cur)
if cur < 0:
cur = 0
if res == 0:
return max(arr)
return res
def soln(n, m, a, b):
poses = []
negs = []
for i in b:
if i < 0:
negs.append(i)
else:
poses.append(i)
return max(kadane(negs + a + poses), kadane(poses + a + negs))
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
print(soln(n, m, a, b)) | FUNC_DEF ASSIGN VAR VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER RETURN FUNC_CALL VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | t = int(input())
i = 0
while t > i:
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
k = []
l = a.copy()
j = 0
while j < m:
if b[j] > 0:
k.append(b[j])
j += 1
a.extend(k)
s = 0
max_till_here = 0
global_max = a[s]
while s < len(a):
max_till_here += a[s]
if max_till_here > global_max:
global_max = max_till_here
if max_till_here <= 0:
max_till_here = 0
s += 1
x = global_max
j = 0
while j < m:
if b[j] > 0:
l.insert(0, b[j])
j += 1
s = 0
max_till_here = 0
global_max = a[s]
while s < len(l):
max_till_here += l[s]
if max_till_here > global_max:
global_max = max_till_here
if max_till_here <= 0:
max_till_here = 0
s += 1
z = global_max
if x > z:
print(x)
else:
print(z)
i += 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | for _ in range(int(input())):
N = int(input())
A = list(map(int, input().split()))
M = int(input())
B = list(map(int, input().split()))
mx = float("-inf")
curmax = 0
curmax2 = 0
stmx = -1
enmx = -1
for i in range(N):
curmax += A[i]
curmax2 += A[-i - 1]
if curmax > mx:
mx = curmax
if i == N - 1:
enmx = curmax
stmx = curmax2
if curmax < 0:
curmax = 0
if curmax2 < 0:
curmax2 = 0
z = sum([x for x in B if x > 0])
print(max(z, z + stmx, z + enmx, mx)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | for t in range(int(input())):
a = int(input())
l = list(map(int, input().split()))
input()
t = list(map(int, input().split()))
k = [l[0]]
m = 0
if k[0] > 0:
m = k[0]
x = 0
y = 0
for i in range(1, a):
if k[-1] > 0:
k += [k[-1] + l[i]]
if k[-1] > m:
m = k[i]
y = i
else:
k += [l[i]]
if x:
x, z = i, x
else:
x = i
z = 0
h = 0
if y != a - 1 and x != 0:
if x < y:
h = sum(l[:x])
elif z != 0:
h = sum(l[:z])
h = max(h, sum(l[y + 1 :]))
s = 0
for i in t:
if i > 0:
s += i
if s + h > 0:
m += s + h
print(m) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR NUMBER NUMBER VAR LIST BIN_OP VAR NUMBER VAR VAR IF VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR LIST VAR VAR IF VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | for _ in range(int(input())):
n = int(input())
ln = list(map(int, input().split()))
m = int(input())
lm = list(map(int, input().split()))
s1 = 0
for i in range(m):
if lm[i] > 0:
s1 += lm[i]
else:
pass
s = ln[0]
z = ln[0]
ind1 = 0
for j in range(1, n):
z = z + ln[j]
if s <= z:
s = z
ind1 = j
s2 = ln[-1]
z2 = ln[-1]
ind2 = 0
for j in range(n - 2, -1, -1):
z2 = z2 + ln[j]
if s2 <= z2:
s2 = z2
ind2 = j
if s2 > s:
s_max = s2 + s1
else:
s_max = s + s1
print(s_max) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | def sumOfMaximumSumSubArray(a):
sm = mx = a[0]
for i in a[1:]:
if sm + i <= i:
sm = 0
sm += i
mx = max(mx, sm)
return mx
def sumOfPositiveElements(a):
res = 0
for i in a:
if i > 0:
res += i
return res
def solve():
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
mx = sumOfPositiveElements(b)
lmx = sumOfMaximumSumSubArray([mx, *a])
rmx = sumOfMaximumSumSubArray([*a, mx])
return max(lmx, rmx)
test_cases_count = int(input())
for tc in range(1, test_cases_count + 1):
result = solve()
print(result)
pass | FUNC_DEF ASSIGN VAR VAR VAR NUMBER FOR VAR VAR NUMBER IF BIN_OP VAR VAR VAR ASSIGN VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR LIST VAR VAR ASSIGN VAR FUNC_CALL VAR LIST VAR VAR RETURN FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | def calc_max_sum_for_array(array):
max_sum = -(10**15)
cur_sum = 0
for el in array:
cur_sum += el
if cur_sum > max_sum:
max_sum = cur_sum
return max_sum
def calc_max_sum(arrayA, arrayB):
max_sum_a = max(
calc_max_sum_for_array(arrayA), calc_max_sum_for_array(arrayA[::-1])
)
max_sum_b = sum([el for el in arrayB if el > 0])
return max_sum_a + max_sum_b
for _ in range(int(input())):
input()
arrayA = list(map(int, input().split()))
input()
arrayB = list(map(int, input().split()))
print(calc_max_sum(arrayA, arrayB)) | FUNC_DEF ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER RETURN BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
c = a.copy()
m = int(input())
b = list(map(int, input().split()))
for i in b:
if i > 0:
a.append(i)
c.insert(0, i)
else:
a.insert(0, i)
c.append(i)
s1 = 0
s2 = 0
ans1 = 0
ans2 = 0
for i in range(m + n):
s1 = s1 + a[i]
s2 = s2 + c[i]
ans1 = max(ans1, s1)
ans2 = max(ans2, s2)
if s1 <= 0:
s1 = 0
if s2 <= 0:
s2 = 0
print(max(ans1, ans2)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | T = int(input())
for i in range(T):
N = int(input())
A = [int(x) for x in input().split()]
M = int(input())
B = [int(x) for x in input().split()]
best_sum = float("-inf")
best_start = best_end = None
current_sum = 0
for current_end, x in enumerate(A):
if current_sum <= 0:
current_start = current_end
current_sum = x
else:
current_sum += x
if current_sum > best_sum:
best_sum = current_sum
best_start = current_start
best_end = current_end
temp = 0
for j in B:
if j > 0:
temp += j
if best_start == 0 or best_end == N - 1:
print(best_sum + temp)
elif sum(A[0:best_start]) >= sum(A[best_end + 1 : N]):
if sum(A[0 : best_end + 1]) + temp > best_sum:
print(sum(A[0 : best_end + 1]) + temp)
else:
print(best_sum)
elif sum(A[best_start:N]) + temp > best_sum:
print(sum(A[best_start:N]) + temp)
else:
print(best_sum) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR NONE ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR IF BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | n = int(input())
for i in range(n):
a = int(input())
b = list(map(int, input().split()))
c = int(input())
d = list(map(int, input().split()))
x = []
y = b[::-1]
z = 0
for i in b:
z += i
x.append(z)
z = 0
for i in y:
z += i
x.append(z)
s = max(x)
for i in d:
if i > 0:
s += i
print(s) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | for _ in range(int(input())):
N = int(input())
A = list(map(int, input().split()))[:N]
s = 0
m = A[0]
for i in range(N):
s += A[i]
if s > m:
m = s
s = 0
for i in range(N - 1, -1, -1):
s += A[i]
if s > m:
m = s
M = int(input())
B = list(map(int, input().split()))[:M]
for i in B:
if i > 0:
m += i
print(m) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FOR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | T = int(input())
for i in range(T):
n = int(input())
a = [int(i) for i in input().split()]
m = int(input())
b = [int(i) for i in input().split()]
def kadane(arr):
max_current = arr[0]
maxi = max_current
for j in arr[1:]:
max_current = max(j, j + max_current)
if max_current > maxi:
maxi = max_current
return maxi
plus = []
minus = []
for k in b:
if k > 0:
plus.append(k)
else:
minus.append(k)
print(max(kadane(a), kadane(plus + a + minus), kadane(minus + a + plus))) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FOR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR RETURN VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | t = int(input())
for j in range(t):
n = int(input())
a = list(map(int, input().split()))
r = int(input())
b = list(map(int, input().split()))
s = a[0]
e = a[0]
for i in range(1, len(a)):
e = max(e + a[i], a[i])
if s < e:
s = e
ans = 0
for k in b:
if k > 0:
ans += k
print(s + ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | for _ in range(int(input())):
msn = int(input())
msa = list(map(int, input().split()))
msm = int(input())
msb = list(map(int, input().split()))
s1 = 0
for i in msb:
if i > 0:
s1 += i
a = s1
m1 = 0
for j in msa:
a = a + j
m1 = max(a, m1)
m2 = 0
a = s1
for u in msa[::-1]:
a += u
m2 = max(m2, a)
print(max(s1, m1, m2)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | def maximumSubarray():
numTestCases = int(input())
for i in range(numTestCases):
sizeA = int(input())
arrayA = [int(x) for x in input().split()]
sizeB = int(input())
arrayB = [int(x) for x in input().split()]
start = 0
end = 0
arrayC = arrayA.copy()
for j in range(sizeB):
if arrayB[j] < 0:
arrayA.append(arrayB[j])
arrayC.insert(0, arrayB[j])
else:
arrayC.append(arrayB[j])
arrayA.insert(0, arrayB[j])
maxA = helper(arrayA, len(arrayA))
maxC = helper(arrayC, len(arrayC))
ansA = 0
sumA = 0
ansC = 0
sumC = 0
for i in range(sizeB + sizeA):
sumA = sumA + arrayA[i]
sumC = sumC + arrayC[i]
ansA = max(sumA, ansA)
ansC = max(sumC, ansC)
if sumA < 0:
sumA = 0
if sumC < 0:
sumC = 0
print(max(maxA, maxC))
def helper(array, length):
maxArray = [0] * length
ans = 0
sum = 0
for i in range(length):
sum = sum + array[i]
ans = max(sum, ans)
if sum < 0:
sum = 0
return ans
maximumSubarray() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | from sys import maxsize
def maxSubArraySum(a, size):
max_so_far = -maxsize - 1
max_ending_here = 0
start = 0
end = 0
s = 0
for i in range(0, size):
max_ending_here += a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
start = s
end = i
if max_ending_here < 0:
max_ending_here = 0
s = i + 1
return [max_so_far, start, end]
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
b.sort(reverse=True)
l = maxSubArraySum(a, n)
su = 0
for x in range(m):
if b[x] > 0:
su += b[x]
else:
break
if l[2] + 1 < n:
su2 = sum(a[l[2] + 1 :])
else:
su2 = 0
print(max(l[0], l[0] + sum(a[: l[1]]) + su, l[0] + su2 + su)) | FUNC_DEF ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN LIST VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | def kadanes(arr):
s = 0
maxi = arr[0]
for i in arr:
s += i
if s > maxi:
maxi = s
if s < 0:
s = 0
return maxi
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
x = int(input())
b = list(map(int, input().split()))
p_dummy = []
n_dummy = []
for i in b:
if i > 0:
p_dummy.append(i)
else:
n_dummy.append(i)
new_sum1 = kadanes(n_dummy + a + p_dummy)
new_sum2 = kadanes(p_dummy + a + n_dummy)
print(max(new_sum1, new_sum2)) | FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | t = int(input())
def kadanes(size, arr):
maxTillNow = 0
maxEnding = 0
for a in arr:
maxEnding += a
if maxEnding < 0:
maxEnding = 0
if maxTillNow < maxEnding:
maxTillNow = maxEnding
return maxTillNow
for _ in range(t):
n = int(input())
leftList = list(map(int, input().split()))
rightList = leftList[:]
m = int(input())
b = list(map(int, input().split()))
for num in b:
if num > 0:
leftList.insert(0, num)
rightList.append(num)
ans = max(kadanes(len(leftList), leftList), kadanes(len(rightList), rightList))
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | for i in range(int(input())):
x = int(input())
arr = list(map(int, input().split()))
y = int(input())
arr2 = list(map(int, input().split()))
m, c1, c2 = 0, 0, 0
for i in range(x):
c1 += arr[i]
c2 += arr[x - i - 1]
m = max(m, max(c1, c2))
c3 = 0
for i in range(y):
c3 += arr2[i] if arr2[i] > 0 else 0
print(m + c3) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Given two arrays A and B of sizes N and M respectively. You can apply the following operation until the array B is non-empty:
Choose either the first or the last element of array B.
Insert the chosen element to either the front or the back of array A.
Delete the chosen element from array B.
For example, let A = [9, 7] and B = [1, 3, 2]. In one operation, we can choose either X = 1 or X = 2 (first or last element of array B). We can insert X in array A and make it either A = [X, 9, 7] or A = [9, 7, X]. The chosen X is deleted from array B. Thus, it will become either B = [3, 2] (when chosen X is 1) or B = [1, 3] (when chosen X is 2).
Find the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
Note: A subarray of an array is formed by deleting some (possibly zero) elements from the beginning of the array and some (possible zero) elements from the end of the array. A subarray can be empty as well.
------ Input Format ------
- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case consists of 4 lines of input.
- The first line of each test contains a single integer N, the size of array A.
- The next line contains N space-separated integers, denoting elements of array A.
- The third line of each test contains a single integer M, the size of array B.
- The next line contains M space-separated integers, denoting elements of array B.
------ Output Format ------
For each test case, output on a new line the maximum sum of any subarray of the array A that you can achieve after performing exactly M operations.
------ Constraints ------
$1 ≤ T ≤ 10$
$1 ≤ N ≤ 10^{5}$
$1 ≤ M ≤ 10^{5}$
$-10^{8} ≤ A_{i}, B_{i} ≤ 10^{8}$
----- Sample Input 1 ------
3
5
3 26 -79 72 23
2
66 44
1
81
1
-97
5
10 -5 14 -20 4
3
-10 5 -2
----- Sample Output 1 ------
205
81
24
----- explanation 1 ------
Test case $1$:
- Operation $1$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66]$ and $B = [44]$.
- Operation $2$: Add the first element of array $B$ to the back of array $A$. Thus, $A = [3, 26, -79, 72, 23, 66, 44]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[72, 23, 66, 44]$ having sum $72+23+66+44=205$.
Test case $2$:
- Operation $1$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-97, 81]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[81]$ having sum $81$.
Test case $3$:
- Operation $1$: Add the last element of array $B$ to the back of array $A$. Thus, $A = [10, -5, 14, -20, 4, -2]$ and $B = [-10, 5]$.
- Operation $2$: Add the last element of array $B$ to the front of array $A$. Thus, $A = [5, 10, -5, 14, -20, 4, -2]$ and $B = [-10]$.
- Operation $3$: Add the first element of array $B$ to the front of array $A$. Thus, $A = [-10, 5, 10, -5, 14, -20, 4, -2]$ and $B = []$.
The, maximum sum subarray of array $A$ is $[5, 10, -5, 14]$ having sum $5+10-5+14 = 24$. | t = int(input())
for iter in range(t):
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
sumOfB = 0
for i in b:
if i > 0:
sumOfB += i
sumOfA = 0
finalMax = a[0]
for i in a:
sumOfA += i
if sumOfA > finalMax:
finalMax = sumOfA
sumOfA = 0
for i in a[::-1]:
sumOfA += i
if sumOfA > finalMax:
finalMax = sumOfA
ans = sumOfB + finalMax
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR NUMBER VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
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