description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
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Our geek loves to play with strings, Currently, he is trying to reduce the size of a string by recursively removing all the consecutive duplicate pairs. In other words, He can apply the below operations any number of times.
Remove all the consecutive duplicate pairs and concatenate the remaining string to replace the original string.
Your task is to find the string with minimum length after applying the above operations.
Note: If the string length become zero after applying operations, return "-1" as a string.
Example 1:
Input:
aaabbaaccd
Output:
ad
Explanation:
Remove (aa)abbaaccd =>abbaaccd
Remove a(bb)aaccd => aaaccd
Remove (aa)accd => accd
Remove a(cc)d => ad
Example 2:
Input:
aaaa
Output:
Empty String
Explanation:
Remove (aa)aa => aa
Again removing pair of duplicates then (aa)
will be removed and we will get 'Empty String'.
Your Task:
This is a function problem. You only need to complete the function removePair() that takes a string as a parameter and returns the modified string. Return "-1" if the whole string is deleted.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 <= |str| <= 10^{4} | class Solution:
def removePair(self, s):
stack = []
ans = ""
for i in range(len(s)):
if stack and stack[-1] == s[i]:
stack.pop()
else:
stack.append(s[i])
if len(stack) == 0:
return -1
else:
while len(stack) > 0:
ans += stack[-1]
stack.pop()
return ans[::-1] | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER WHILE FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR RETURN VAR NUMBER |
Our geek loves to play with strings, Currently, he is trying to reduce the size of a string by recursively removing all the consecutive duplicate pairs. In other words, He can apply the below operations any number of times.
Remove all the consecutive duplicate pairs and concatenate the remaining string to replace the original string.
Your task is to find the string with minimum length after applying the above operations.
Note: If the string length become zero after applying operations, return "-1" as a string.
Example 1:
Input:
aaabbaaccd
Output:
ad
Explanation:
Remove (aa)abbaaccd =>abbaaccd
Remove a(bb)aaccd => aaaccd
Remove (aa)accd => accd
Remove a(cc)d => ad
Example 2:
Input:
aaaa
Output:
Empty String
Explanation:
Remove (aa)aa => aa
Again removing pair of duplicates then (aa)
will be removed and we will get 'Empty String'.
Your Task:
This is a function problem. You only need to complete the function removePair() that takes a string as a parameter and returns the modified string. Return "-1" if the whole string is deleted.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 <= |str| <= 10^{4} | class Solution:
def removePair(self, s):
li = []
lis = list(s)
for item in lis:
if li and li[-1] == item:
li.pop()
else:
li.append(item)
result = "".join(li)
if result:
return result
return -1 | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR IF VAR VAR NUMBER VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING VAR IF VAR RETURN VAR RETURN NUMBER |
Our geek loves to play with strings, Currently, he is trying to reduce the size of a string by recursively removing all the consecutive duplicate pairs. In other words, He can apply the below operations any number of times.
Remove all the consecutive duplicate pairs and concatenate the remaining string to replace the original string.
Your task is to find the string with minimum length after applying the above operations.
Note: If the string length become zero after applying operations, return "-1" as a string.
Example 1:
Input:
aaabbaaccd
Output:
ad
Explanation:
Remove (aa)abbaaccd =>abbaaccd
Remove a(bb)aaccd => aaaccd
Remove (aa)accd => accd
Remove a(cc)d => ad
Example 2:
Input:
aaaa
Output:
Empty String
Explanation:
Remove (aa)aa => aa
Again removing pair of duplicates then (aa)
will be removed and we will get 'Empty String'.
Your Task:
This is a function problem. You only need to complete the function removePair() that takes a string as a parameter and returns the modified string. Return "-1" if the whole string is deleted.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 <= |str| <= 10^{4} | class Solution:
def removePair(self, S):
s = []
for i in S:
if s and s[-1] == i:
s.pop()
else:
s.append(i)
if len(s) == 0:
return -1
else:
return "".join(s) | CLASS_DEF FUNC_DEF ASSIGN VAR LIST FOR VAR VAR IF VAR VAR NUMBER VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER RETURN FUNC_CALL STRING VAR |
Our geek loves to play with strings, Currently, he is trying to reduce the size of a string by recursively removing all the consecutive duplicate pairs. In other words, He can apply the below operations any number of times.
Remove all the consecutive duplicate pairs and concatenate the remaining string to replace the original string.
Your task is to find the string with minimum length after applying the above operations.
Note: If the string length become zero after applying operations, return "-1" as a string.
Example 1:
Input:
aaabbaaccd
Output:
ad
Explanation:
Remove (aa)abbaaccd =>abbaaccd
Remove a(bb)aaccd => aaaccd
Remove (aa)accd => accd
Remove a(cc)d => ad
Example 2:
Input:
aaaa
Output:
Empty String
Explanation:
Remove (aa)aa => aa
Again removing pair of duplicates then (aa)
will be removed and we will get 'Empty String'.
Your Task:
This is a function problem. You only need to complete the function removePair() that takes a string as a parameter and returns the modified string. Return "-1" if the whole string is deleted.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 <= |str| <= 10^{4} | class Solution:
def removePair(self, s):
l1 = []
for i in range(len(s)):
if l1 == []:
l1.append(s[i])
elif l1[-1] == s[i]:
l1.pop()
else:
l1.append(s[i])
s1 = "".join(l1)
if s1 == "":
return -1
else:
return s1 | CLASS_DEF FUNC_DEF ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR LIST EXPR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL STRING VAR IF VAR STRING RETURN NUMBER RETURN VAR |
Our geek loves to play with strings, Currently, he is trying to reduce the size of a string by recursively removing all the consecutive duplicate pairs. In other words, He can apply the below operations any number of times.
Remove all the consecutive duplicate pairs and concatenate the remaining string to replace the original string.
Your task is to find the string with minimum length after applying the above operations.
Note: If the string length become zero after applying operations, return "-1" as a string.
Example 1:
Input:
aaabbaaccd
Output:
ad
Explanation:
Remove (aa)abbaaccd =>abbaaccd
Remove a(bb)aaccd => aaaccd
Remove (aa)accd => accd
Remove a(cc)d => ad
Example 2:
Input:
aaaa
Output:
Empty String
Explanation:
Remove (aa)aa => aa
Again removing pair of duplicates then (aa)
will be removed and we will get 'Empty String'.
Your Task:
This is a function problem. You only need to complete the function removePair() that takes a string as a parameter and returns the modified string. Return "-1" if the whole string is deleted.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 <= |str| <= 10^{4} | class Solution:
def removePair(self, s):
li = list(s)
i = 0
while i < len(li) - 1:
if li[i] == li[i + 1]:
li.pop(i + 1)
li.pop(i)
if i != 0:
i = i - 1
else:
i = i + 1
if len(li) != 0:
return "".join(li)
else:
return -1 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL STRING VAR RETURN NUMBER |
Our geek loves to play with strings, Currently, he is trying to reduce the size of a string by recursively removing all the consecutive duplicate pairs. In other words, He can apply the below operations any number of times.
Remove all the consecutive duplicate pairs and concatenate the remaining string to replace the original string.
Your task is to find the string with minimum length after applying the above operations.
Note: If the string length become zero after applying operations, return "-1" as a string.
Example 1:
Input:
aaabbaaccd
Output:
ad
Explanation:
Remove (aa)abbaaccd =>abbaaccd
Remove a(bb)aaccd => aaaccd
Remove (aa)accd => accd
Remove a(cc)d => ad
Example 2:
Input:
aaaa
Output:
Empty String
Explanation:
Remove (aa)aa => aa
Again removing pair of duplicates then (aa)
will be removed and we will get 'Empty String'.
Your Task:
This is a function problem. You only need to complete the function removePair() that takes a string as a parameter and returns the modified string. Return "-1" if the whole string is deleted.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 <= |str| <= 10^{4} | class Solution:
def removePair(self, s):
d = []
p = ""
for i in range(len(s)):
if len(d) == 0:
d.append(s[i])
elif d[-1] != s[i]:
d.append(s[i])
elif d[-1] == s[i]:
d.pop()
for j in d:
p += j
if p:
return p
else:
return -1 | CLASS_DEF FUNC_DEF ASSIGN VAR LIST ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR FOR VAR VAR VAR VAR IF VAR RETURN VAR RETURN NUMBER |
Our geek loves to play with strings, Currently, he is trying to reduce the size of a string by recursively removing all the consecutive duplicate pairs. In other words, He can apply the below operations any number of times.
Remove all the consecutive duplicate pairs and concatenate the remaining string to replace the original string.
Your task is to find the string with minimum length after applying the above operations.
Note: If the string length become zero after applying operations, return "-1" as a string.
Example 1:
Input:
aaabbaaccd
Output:
ad
Explanation:
Remove (aa)abbaaccd =>abbaaccd
Remove a(bb)aaccd => aaaccd
Remove (aa)accd => accd
Remove a(cc)d => ad
Example 2:
Input:
aaaa
Output:
Empty String
Explanation:
Remove (aa)aa => aa
Again removing pair of duplicates then (aa)
will be removed and we will get 'Empty String'.
Your Task:
This is a function problem. You only need to complete the function removePair() that takes a string as a parameter and returns the modified string. Return "-1" if the whole string is deleted.
Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)
Constraints:
1 <= |str| <= 10^{4} | class Solution:
def removePair(self, s):
dup_index = self.checkDuplicates(s)
while dup_index < len(s) - 1:
s = s[0:dup_index] + s[dup_index + 2 :]
dup_index = self.checkDuplicates(s)
if len(s) > 0:
return s
else:
return -1
def checkDuplicates(self, s):
i = 0
for i in range(0, len(s) - 1):
if s[i] == s[i + 1]:
return i
return i + 1 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN VAR RETURN NUMBER FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER RETURN VAR RETURN BIN_OP VAR NUMBER |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | T = int(input())
for i in range(T):
s = input()
x = 0
y = len(s) - 1
z = 1
while x <= y:
if s[x] == "?" == s[y]:
z = z * 26 % 10000009
elif s[x] != "?" and s[y] != "?" and s[x] != s[y]:
z = 0
break
x += 1
y -= 1
print(z) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR VAR STRING VAR VAR VAR VAR ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | MOD = 10000009
def solve():
s = input()
ls = len(s)
res = 1
if ls == 1:
if s[0] == "?":
print(26)
else:
print(1)
return
if ls & 1:
for i in range(ls // 2):
first_ch = s[i]
last_ch = s[ls - i - 1]
if first_ch == "?" and last_ch == "?":
res = res * 26 % MOD
elif first_ch == "?" or last_ch == "?":
res = res * 1 % MOD
elif first_ch != last_ch:
res = res * 0 % MOD
if s[ls // 2] == "?":
res = res * 26 % MOD
else:
for i in range(ls // 2):
first_ch = s[i]
last_ch = s[ls - i - 1]
if first_ch == "?" and last_ch == "?":
res = res * 26 % MOD
elif first_ch == "?" or last_ch == "?":
res = res * 1 % MOD
elif first_ch != last_ch:
res = res * 0 % MOD
print(res)
for _ in range(int(input())):
solve() | ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN IF BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR STRING VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR STRING VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR STRING VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR STRING VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | t = int(input())
for _ in range(t):
s = input()
n = len(s)
if n == 1 and s[0] == "?":
print(26)
else:
count = 1
left = 0
right = n - 1
while left <= right:
if s[left] == "?" and s[right] == "?":
left += 1
right -= 1
count = count * 26 % 10000009
elif s[left] == "?" or s[right] == "?" or s[left] == s[right]:
left += 1
right -= 1
continue
else:
count = 0
break
print(count) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR VAR STRING VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | mod = 10000009
for _ in range(int(input())):
st = input()
c = 1
i = 0
j = len(st) - 1
while i <= j:
if st[i] == "?" and st[j] == "?":
c = c * 26 % mod
elif st[i] == "?" or st[j] == "?":
pass
elif st[i] != st[j]:
c = 0
break
i = i + 1
j = j - 1
print(c % mod) | ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR VAR STRING VAR VAR STRING IF VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | t = int(input())
mod = 10000009
while t:
s = input()
m = len(s)
ans = 1
for i in range(m // 2 + 1):
if i == m // 2 and m & 1 == 0:
break
if (s[i] != "?" and s[-(i + 1)] != "?") and s[i] != s[-(i + 1)]:
ans = 0
break
if s[i] == s[-(i + 1)] == "?":
ans = ans * 26 % mod
print(ans % mod)
t -= 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | k = 10**7 + 9
for _ in range(int(input())):
s = input()
n = len(s)
i, j = 0, n - 1
c, f = 1, 0
while i <= j:
if s[i] == s[j]:
if s[i] == "?":
c = c % k * 26 % k
elif s[i] != s[j]:
if s[i] == "?" or s[j] == "?":
pass
else:
c = 0
break
i += 1
j -= 1
print(c) | ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER WHILE VAR VAR IF VAR VAR VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR IF VAR VAR VAR VAR IF VAR VAR STRING VAR VAR STRING ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
s1 = str(input())
n = len(s1)
ans = 1
for i in range((n + 1) // 2):
if s1[i] == "?" and s1[n - i - 1] == "?":
ans = ans * 26 % 10000009
elif s1[i] != "?" and s1[n - i - 1] != "?" and s1[i] != s1[n - i - 1]:
ans = 0
break
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for j in range(int(input())):
s = input()
l = len(s)
ans = 1
mod = 10000009
for i in range((l + 1) // 2):
if s[i] != s[l - 1 - i] and s[i] != "?" and s[l - 1 - i] != "?":
ans = 0
break
elif s[i] == "?" and s[l - 1 - i] == "?":
ans = ans * 26 % mod
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR VAR STRING VAR BIN_OP BIN_OP VAR NUMBER VAR STRING ASSIGN VAR NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR NUMBER VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for u in range(int(input())):
s = input()
i = 0
j = len(s) - 1
c = 1
m = 10**7 + 9
while i <= j:
if s[i] == s[j] == "?":
c = 26 % m * (c % m) % m
i += 1
j -= 1
elif s[i] == "?" or s[j] == "?" or s[i] == s[j]:
i += 1
j -= 1
else:
c = 0
break
print(c % m) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER WHILE VAR VAR IF VAR VAR VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR BIN_OP VAR VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR STRING VAR VAR STRING VAR VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | from sys import stdin, stdout
mod = 10**7 + 9
for _ in range(int(stdin.readline())):
s = input().strip()
n = len(s)
m = 1
for i in range(n // 2):
if s[i] == s[n - i - 1] and s[i] == "?":
m = m * 26 % mod
elif s[i] != s[n - i - 1] and (s[i] == "?" or s[n - i - 1] == "?"):
pass
elif s[i] == s[n - i - 1]:
pass
else:
print(0)
break
else:
if n % 2 == 1 and s[n // 2] == "?":
m = m * 26 % mod
print(m) | ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER 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 NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | tests = int(input())
for i in range(tests):
string = input()
length = len(string)
ans = 1
end = (length - 1) // 2 + 1
for j in range(end):
if string[j] == "?" and string[-j - 1] == "?":
ans *= 26
elif string[j] == "?" or string[-j - 1] == "?":
ans *= 1
elif string[j] != string[-j - 1]:
ans = 0
break
ans %= 10000009
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | t = int(input())
for i in range(t):
s = input()
c = 1
d = len(s)
for i in range((d + 1) // 2):
j = d - 1 - i
if s[i] != "?" and s[j] != "?" and s[i] != s[j]:
c = 0
break
elif s[i] == "?" and s[j] == "?":
c = c * 26 % 10000009
print(c) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR VAR STRING VAR VAR STRING VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR STRING VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | MOD = 10000009
def count_palindromes(s):
n = len(s)
count = 1
for i in range(n // 2):
j = n - i - 1
if s[i] == "?" and s[j] == "?":
count = count * 26 % MOD
elif s[i] == s[j] or s[i] == "?" or s[j] == "?":
continue
else:
return 0
if n % 2 == 1 and s[n // 2] == "?":
count = count * 26 % MOD
return count
t = int(input())
for _ in range(t):
s = input().strip()
print(count_palindromes(s)) | ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR STRING VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR VAR VAR VAR VAR VAR STRING VAR VAR STRING RETURN NUMBER IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | t = 0
try:
t = int(input())
except:
pass
for _ in range(t):
l = input()
n = len(l)
prod = 1
for k in range(n // 2):
i = l[k]
j = l[n - k - 1]
if i != j and (i != "?" and j != "?"):
prod = 0
break
elif i == j and i == "?":
prod *= 26
prod = prod % 10000009
if n % 2 != 0:
if l[n // 2] == "?":
prod *= 26
print(prod) | ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR STRING VAR STRING ASSIGN VAR NUMBER IF VAR VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for t in range(int(input())):
s = input()
n = len(s)
c, k = 1, 0
for i in range((len(s) - 1) // 2 + 1):
if s[i] == "?" and s[n - i - 1] == "?":
c = c * 26 % 10000009
elif s[i] == "?" or s[n - i - 1] == "?":
pass
elif s[i] != s[n - i - 1]:
k = 1
break
if k == 1:
print(0)
else:
print(c) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | mod = 10000009
t = int(input())
for _ in range(t):
s = input()
ln = len(s)
dx = 1
for i in range(ln // 2):
xn = ln - i - 1
x = s[i]
y = s[xn]
if x == "?" and y == "?":
dx = dx * 26 % mod
elif x == y:
continue
elif x == "?" or y == "?":
continue
else:
dx = 0
break
if ln % 2 == 1 and ln > 1:
if s[ln // 2] == "?":
dx = dx * 26 % mod
elif ln == 1:
if s == "?":
dx = 26 % mod
print(dx) | ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF VAR STRING VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR VAR IF VAR STRING VAR STRING ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR STRING ASSIGN VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
str1 = str(input())
n = len(str1)
a, b, c = 0, n - 1, 1
while a <= b:
if str1[a] == "?" == str1[b]:
c = c * 26 % 10000009
elif str1[a] != "?" and str1[b] != "?" and str1[a] != str1[b]:
c = 0
break
a += 1
b -= 1
print(c) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER WHILE VAR VAR IF VAR VAR STRING VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR VAR STRING VAR VAR VAR VAR ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
s = input()
nWays = 1
n = len(s)
for i in range(n // 2):
if s[i] == "?" and s[-1 - i] == "?":
nWays = nWays * 26 % 10000009
elif s[i] != "?" and s[-1 - i] != "?" and s[i] != s[-i - 1]:
nWays = 0
break
if n % 2 and s[n // 2] == "?":
nWays = nWays * 26 % (10**7 + 9)
print(nWays) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR BIN_OP NUMBER VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP NUMBER VAR STRING VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
pal = input().strip()
mul = 0
flag = True
if len(pal) % 2 == 0:
till = len(pal) // 2
else:
till = len(pal) // 2 + 1
for i in range(till):
left, right = pal[i], pal[len(pal) - i - 1]
if left == "?":
if right == "?":
mul += 1
elif right == "?":
continue
elif left != right:
flag = False
break
if flag:
print(max(26**mul, 1) % 10000009)
else:
print("0") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR STRING IF VAR STRING VAR NUMBER IF VAR STRING IF VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP NUMBER VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | a = int(input())
for _ in range(a):
b = list(input())
s = 1
for i in range(len(b) // 2):
if b[i] == "?" and b[len(b) - 1 - i] == "?":
s = s * 26 % 10000009
elif b[i] == "?" or b[len(b) - 1 - i] == "?":
s *= 1
elif b[i] != b[len(b) - 1 - i]:
s = 0
c = True
break
if len(b) % 2 == 1:
if b[len(b) // 2] == "?":
s = s * 26 % 10000009
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 NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR STRING VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | mod = 10000009
def solve():
s = input()
n = len(s)
c, k = 1, 0
for i in range((len(s) - 1) // 2 + 1):
if s[i] == "?" and s[n - i - 1] == "?":
c = c * 26 % 10000009
elif s[i] == "?" or s[n - i - 1] == "?":
pass
elif s[i] != s[n - i - 1]:
k = 1
break
if k == 1:
print(0)
else:
print(c)
def __starting_point():
t = int(input())
while t != 0:
solve()
t -= 1
__starting_point() | ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | t = int(input())
while t > 0:
t -= 1
s = input()
n = len(s)
l = 0
r = n - 1
count = 0
imposible = False
while l <= r:
if s[l] == "?" and s[r] == "?":
count += 1
elif s[l] != "?" and s[r] != "?" and s[l] != s[r]:
imposible = True
l += 1
r -= 1
if imposible == True:
print("0")
else:
ans = 1
for i in range(count):
ans = ans * 26 % 10000009
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR VAR STRING VAR NUMBER IF VAR VAR STRING VAR VAR STRING VAR VAR VAR VAR ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
s = input()
n = len(s)
count = 0
for i in range(n // 2):
if s[i] != s[n - i - 1]:
if s[i] != "?" and s[n - 1 - i] != "?":
print(0)
break
elif s[i] == "?" and s[n - i - 1] == "?":
count += 1
else:
if n & 1 == 1:
if s[n // 2] == "?":
count += 1
print(pow(26, count, 10000009)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR NUMBER VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR NUMBER |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
s = list(input())
n = len(s)
an = 1
mod = 10000009
for i in range(n):
if s[i] == "?" and s[n - i - 1] == "?":
an = an * 26 % mod
s[i] = "."
s[n - i - 1] = "."
elif s[i] != "?" and s[n - i - 1] != "?":
if s[i] != s[n - i - 1]:
an = 0
break
print(an % mod) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR ASSIGN VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER STRING IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
s = input()
i = 0
j = len(s) - 1
cnt = 1
f = 1
while i <= j:
if s[i] == "?" and s[j] == "?":
cnt = cnt * 26 % 10000009
elif s[i] == "?" or s[j] == "?":
pass
elif s[i] != s[j]:
f = -1
break
i += 1
j -= 1
if f == -1:
print(0)
else:
print(cnt) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR STRING VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR VAR STRING IF VAR VAR VAR VAR ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
s = input()
l = len(s)
flag = False
count = 0
if l % 2 == 1:
if s[l // 2] == "?":
count += 1
for i in range(0, l // 2):
if s[i] == "?" and s[l - i - 1] == "?":
count += 1
elif s[i] != s[l - i - 1] and s[i] != "?" and s[l - i - 1] != "?":
flag = True
break
if flag:
print(0)
else:
ans = 1
for i in range(count):
ans = ans * 26 % 10000009
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | tc = int(input())
for _ in range(tc):
s = input()
n = len(s)
ans = 1
for i in range(n // 2):
x = (s[i] == "?") + (s[n - i - 1] == "?")
if x == 2:
ans = ans * 26 % 10000009
elif x == 0 and s[i] != s[n - i - 1]:
ans = 0
break
if n % 2 == 1 and s[n // 2] == "?":
ans *= 26
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING IF VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR NUMBER VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER STRING VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | t = int(input())
for _ in range(t):
s = input()
s = [i for i in s]
n = len(s)
c = 1
mc = 1
for i in range(len(s)):
if s[i] == "?" and s[abs(n - 1 - i)] == "?":
s[abs(n - 1 - i)] = s[i] = "a"
c *= 26
elif s[i] != s[abs(n - 1 - i)] and s[i] != "?" and s[abs(n - 1 - i)] != "?":
c = 0
mc = 0
break
c = c % 10000009
print(c) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR VAR STRING VAR NUMBER IF VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR VAR VAR STRING VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | def solve():
s = input()
n = len(s)
c = 1
for i in range((n - 1) // 2 + 1):
if s[i] == "?" and s[n - i - 1] == "?":
c = c * 26 % 10000009
elif s[i] == "?" or s[n - i - 1] == "?":
c = c * 1
elif s[i] != s[n - i - 1]:
c = 0
break
print(c)
t = int(input())
while t > 0:
solve()
t = t - 1 | FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP BIN_OP VAR VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for _ in range(int(input())):
s = input()
nWays = 1
n = len(s)
f = True
for i in range(n // 2):
if s[i] == s[-i - 1]:
if s[i] == "?":
nWays = nWays * 26 % (10**7 + 9)
elif s[i] == "?" and s[-i - 1] != "?" or s[i] != "?" and s[-i - 1] == "?":
continue
else:
nWays = 0
f = False
break
if f and n % 2 and s[n // 2] == "?":
nWays = nWays * 26 % (10**7 + 9)
print(nWays) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR VAR STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | test = int(input())
for _ in range(test):
strText = input()
n = len(strText)
ans = 1
rangeCheck = n
if n % 2 == 0:
rangeCheck = rangeCheck // 2
else:
rangeCheck = rangeCheck // 2 + 1
if len(strText) == strText.count("?"):
ans = ans * 26**rangeCheck % (10**7 + 9)
else:
for i in range(rangeCheck):
start = strText[i]
end = strText[n - 1 - i]
if start == "?" and end == "?":
ans = ans * 26 % (10**7 + 9)
elif start != end:
if start != "?" and end != "?":
ans = 0
break
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP NUMBER VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR NUMBER VAR IF VAR STRING VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER IF VAR VAR IF VAR STRING VAR STRING ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR |
Chef had an interesting dream last night. He dreamed of a new revolutionary chicken recipe. When he woke up today he tried very hard to reconstruct the ingredient list. But, he could only remember certain ingredients. To simplify the problem, the ingredient list can be represented by a string of lowercase characters 'a' - 'z'.
Chef can recall some characters of the ingredient list, all the others, he has forgotten. However, he is quite sure that the ingredient list was a palindrome.
You are given the ingredient list Chef dreamed last night. The forgotten characters are represented by a question mark ('?'). Count the number of ways Chef can replace the forgotten characters with characters 'a' - 'z' in such a way that resulting ingredient list is a palindrome.
-----Input-----
The first line of input contains a single integer T, the number of test cases. T lines follow, each containing a single non-empty string - the ingredient list as recalled by Chef. Whatever letters he couldn't recall are represented by a '?'.
-----Output-----
For each test case, output a single line containing the number of valid ways the ingredient list could be completed. Since the answers can be very large, output each answer modulo 10,000,009.
-----Example-----
Input:
5
?
??
ab?
a?c
aba
Output:
26
26
1
0
1
-----Constraints-----
1 ≤ T ≤ 20
1 ≤ sum of length of all input strings ≤ 1,000,000
Each input string contains only lowercase roman letters ('a' - 'z') or question marks. | for t in range(int(input())):
s = input()
ans = 1
b = 0
du = len(s)
if du == 1:
if s[0] == "?":
print(26)
else:
print(1)
else:
for i in range(du // 2):
if s[i] == "?" and s[-i - 1] == "?":
ans = ans * 26 % 10000009
elif (s[i] == "?" or s[-i - 1] == "?") or s[i] == s[-i - 1]:
ans = ans
elif s[i] != s[-i - 1]:
ans = 0
b = 1
print(ans % 10000009)
break
if du % 2 != 0:
if s[du // 2] == "?":
ans = ans * 26 % 10000009
if b == 0:
print(ans % 10000009) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
total_candies = 0
open_boxes = [box for box in initialBoxes if status[box] == 1]
locked_boxes = [box for box in initialBoxes if status[box] == 0]
obtained_keys = set([])
while len(open_boxes):
box = open_boxes.pop()
total_candies += candies[box]
for key in keys[box]:
if key in locked_boxes:
locked_boxes.remove(key)
open_boxes.append(key)
else:
obtained_keys.add(key)
for cbox in containedBoxes[box]:
if status[cbox] == 1 or cbox in obtained_keys:
open_boxes.append(cbox)
else:
locked_boxes.append(cbox)
return total_candies | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR LIST WHILE FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
currkeys, ret, taken = set(), 0, [0] * len(status)
q = collections.deque(initialBoxes)
while q:
l = len(q)
opened = False
for _ in range(l):
curr = q.popleft()
taken[curr] = 1
if curr not in currkeys and status[curr] == 0:
q.append(curr)
else:
opened = True
ret += candies[curr]
if keys[curr]:
currkeys |= set([key for key in keys[curr]])
for b in containedBoxes[curr]:
if not taken[b]:
q.append(b)
if not opened:
break
return ret | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR VAR VAR IF VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
open_box = []
close_box = set()
key_set = set([i for i in range(len(status)) if status[i] == 1])
visited = [(False) for _ in range(len(status))]
count = 0
for box in initialBoxes:
if status[box] == 1:
open_box.append(box)
else:
close_box.add(box)
while open_box:
box = open_box.pop()
visited[box] = True
count += candies[box]
for key in keys[box]:
key_set.add(key)
for c_box in containedBoxes[box]:
if visited[c_box]:
continue
else:
close_box.add(c_box)
for key in key_set:
if key in close_box:
open_box.append(key)
close_box.remove(key)
return count | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER VAR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
q = initialBoxes
has_key = [0] * len(status)
has_open = [0] * len(status)
res = 0
while q:
qq = []
for s in q:
if status[s] or has_key[s]:
res += candies[s]
has_open[s] = 1
for b in containedBoxes[s]:
if has_open[b]:
continue
qq.append(b)
for k in keys[s]:
has_key[k] = 1
else:
qq.append(s)
if tuple(qq) == tuple(q):
break
q, qq = qq, []
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN 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 WHILE VAR ASSIGN VAR LIST FOR VAR VAR IF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR LIST RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
candy_count = 0
my_boxes = initialBoxes
my_keys = []
while my_boxes:
l = len(my_boxes)
opened_new = False
for b_i in range(l - 1, -1, -1):
my_box = my_boxes[b_i]
if status[my_box] or my_box in my_keys:
my_boxes.pop(b_i)
my_boxes.extend(containedBoxes[my_box])
candy_count += candies[my_box]
my_keys.extend(keys[my_box])
opened_new = True
if not opened_new:
return candy_count
return candy_count | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR LIST WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF VAR RETURN VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Box:
def __init__(self, status, candies, keys, boxes):
self.status = status
self.candies = candies
self.keys = keys
self.containedBoxes = boxes
def openboxes(unopened, keys, Q):
for u in unopened:
if u in keys:
Q.append(u)
unopened.remove(u)
class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
mykeys = []
unopened = []
candySum = 0
Q = []
for i in initialBoxes:
mykeys.extend(keys[i])
if status[i] == 1:
Q.append(i)
else:
unopened.append(i)
openboxes(unopened, mykeys, Q)
while Q != []:
box = Q.pop(0)
candySum += candies[box]
for b in containedBoxes[box]:
mykeys.extend(keys[b])
if status[b] == 1:
Q.append(b)
else:
unopened.append(b)
openboxes(unopened, mykeys, Q)
return candySum | CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR FUNC_DEF FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR WHILE VAR LIST ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
boxes = set(initialBoxes)
queue = collections.deque()
for box in boxes:
if status[box] == 1:
queue.append(box)
ans = 0
seen = set()
while queue:
box = queue.popleft()
if box not in seen:
seen.add(box)
ans += candies[box]
for i in containedBoxes[box]:
boxes.add(i)
if status[i] == 1:
queue.append(i)
for j in keys[box]:
if status[j] == 0 and j in boxes:
queue.append(j)
status[j] = 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
key_set = set()
box_set = set()
used = set()
queue = collections.deque(initialBoxes)
ans = 0
while queue:
box = queue.popleft()
if box in used:
continue
if status[box] == 1 or box in key_set:
used.add(box)
if box in key_set:
key_set.remove(box)
if box in box_set:
box_set.remove(box)
ans += candies[box]
for futureBox in containedBoxes[box]:
box_set.add(futureBox)
for futureKey in keys[box]:
key_set.add(futureKey)
for futureBox in box_set:
if futureBox in key_set or status[futureBox] == 1:
queue.append(futureBox)
else:
box_set.add(box)
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR IF VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
N = len(status)
scores = [0] * N
for i in range(N):
scores[i] += status[i]
level = []
for b in initialBoxes:
scores[b] += N + 1
if scores[b] >= N + 2:
level.append(b)
scores[b] = -sys.maxsize
res = 0
while level:
n_level = []
for b in level:
res += candies[b]
for k in containedBoxes[b]:
scores[k] += N + 1
if scores[k] >= N + 2:
n_level.append(k)
scores[k] = -sys.maxsize
for k in keys[b]:
scores[k] += 1
if scores[k] >= N + 2:
n_level.append(k)
scores[k] = -sys.maxsize
level = n_level
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST FOR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR LIST FOR VAR VAR VAR VAR VAR FOR VAR VAR VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FOR VAR VAR VAR VAR VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
boxes = set(initialBoxes)
openedBox = set()
bfs = [i for i in boxes if status[i]]
while bfs:
size = len(bfs)
for _ in range(size):
box = bfs.pop(0)
openedBox.add(box)
for eachContainedBox in containedBoxes[box]:
if status[eachContainedBox]:
bfs.append(eachContainedBox)
boxes.add(eachContainedBox)
for eachKey in keys[box]:
if status[eachKey] == 0 and eachKey in boxes:
bfs.append(eachKey)
status[eachKey] = 1
return sum(candies[box] for box in openedBox) | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
reachable = [False] * len(status)
visited = [False] * len(status)
for box in initialBoxes:
reachable[box] = True
for i in range(len(containedBoxes)):
for inside in containedBoxes[i]:
reachable[inside] = False
queue = initialBoxes
target = []
ret = 0
while queue:
for box in queue:
if status[box] == 1 and reachable[box] and not visited[box]:
ret += candies[box]
visited[box] = True
for key in keys[box]:
if status[key] == 0:
status[key] = 1
if reachable[key]:
target.append(key)
for inside in containedBoxes[box]:
reachable[inside] = True
if status[inside] == 1:
target.append(inside)
else:
target.append(box)
if target == queue:
break
queue = target
target = []
return ret | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR FOR VAR VAR IF VAR VAR NUMBER VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR LIST RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
s = [b for b in initialBoxes if status[b]]
found_boxes = set([b for b in initialBoxes if not status[b]])
r = 0
while s:
box = s.pop()
r += candies[box]
for key in keys[box]:
status[key] = 1
tbr = set()
for b in found_boxes:
if status[b]:
s.append(b)
tbr.add(b)
found_boxes -= tbr
for child in containedBoxes[box]:
if status[child]:
s.append(child)
else:
found_boxes.add(child)
return r | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
res, bcache, kcache = 0, set(initialBoxes), set()
for _ in range(1000):
tmp = set()
for b in bcache:
if status[b] == 1 or b in kcache:
tmp |= set(containedBoxes[b])
kcache |= set(keys[b])
res += candies[b]
else:
tmp.add(b)
bcache = tmp
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
OPEN, CLOSED = 1, 0
q = deque(initialBoxes)
keys_obtained = set(i for i in range(len(status)) if status[i] == OPEN)
count = 0
while set(q) & keys_obtained:
q2 = deque()
while q:
box = q.popleft()
if box in keys_obtained:
count += candies[box]
keys_obtained |= set(keys[box])
q.extend(containedBoxes[box])
else:
q2.append(box)
q = q2
return count | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
reached = set()
visited = set()
q = initialBoxes
yummy = 0
while q:
new_q = []
for box in q:
reached.add(box)
if status[box] and box not in visited:
visited.add(box)
new_q.extend(containedBoxes[box])
new_q.extend(
unlockedBox
for unlockedBox in keys[box]
if unlockedBox in reached
)
yummy += candies[box]
for unlockedBox in keys[box]:
status[unlockedBox] = 1
q = new_q
return yummy | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
reachable = [(False) for _ in status]
for i in initialBoxes:
reachable[i] = True
can_open = [(val == 1) for val in status]
nodes_to_visit = [b for b in range(len(status)) if reachable[b] and can_open[b]]
total_candies = 0
while len(nodes_to_visit) >= 1:
curr_node = nodes_to_visit[0]
nodes_to_visit = nodes_to_visit[1:]
total_candies += candies[curr_node]
for k in keys[curr_node]:
if not can_open[k] and reachable[k]:
nodes_to_visit.append(k)
can_open[k] = True
for b in containedBoxes[curr_node]:
if not reachable[b] and can_open[b]:
nodes_to_visit.append(b)
reachable[b] = True
return total_candies | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER VAR VAR FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
ret = 0
n = len(status)
available_to_open = [box for box in initialBoxes if status[box]]
reachable_boxes = set([box for box in initialBoxes if not status[box]])
while len(available_to_open) > 0:
box = available_to_open.pop()
if status[box] == 2:
continue
ret += candies[box]
status[box] = 2
for containedBox in containedBoxes[box]:
reachable_boxes.add(containedBox)
for key in keys[box]:
if status[key] == 0:
status[key] = 1
for containedBox in containedBoxes[box]:
if status[containedBox] == 1:
available_to_open.append(containedBox)
for key in keys[box]:
if key in reachable_boxes:
available_to_open.append(key)
return ret
return ret | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR IF VAR VAR NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
if not initialBoxes:
return 0
visited = set()
boxes = set()
keys_found = set()
n_candies = 0
opn = []
for i in initialBoxes:
if status[i] != 1:
boxes.add(i)
else:
opn.append(i)
initialBoxes = opn
while initialBoxes:
i = initialBoxes.pop(0)
n_candies += candies[i]
visited.add(i)
if keys[i] != []:
for j in keys[i]:
keys_found.add(j)
if containedBoxes[i] != []:
for c in containedBoxes[i]:
if status[c] == 1:
initialBoxes.append(c)
else:
boxes.add(c)
for k in boxes:
if k in visited:
continue
if k in keys_found:
initialBoxes.append(k)
visited.add(k)
return n_candies | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR LIST FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR LIST FOR VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR IF VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
n = len(status)
visited = [0] * n
unopened = [0] * n
opened = []
ans = 0
for b in initialBoxes:
if status[b] == 1:
opened.append(b)
visited[b] = 1
else:
unopened[b] = 1
while opened:
q2 = []
for b in opened:
ans += candies[b]
for bb in containedBoxes[b]:
if visited[bb] == 0:
if status[bb] == 1:
q2.append(bb)
else:
unopened[bb] = 1
for k in keys[b]:
status[k] = 1
if unopened[k] == 1 and visited[k] == 0:
unopened[k] = 0
visited[k] = 1
q2.append(k)
opened = q2
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER WHILE VAR ASSIGN VAR LIST FOR VAR VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
foundBoxes = set(initialBoxes)
openBoxes = [box for box in foundBoxes if status[box] == 1]
for i in openBoxes:
for x in containedBoxes[i]:
foundBoxes.add(x)
if status[x] == 1:
openBoxes.append(x)
for x in keys[i]:
if status[x] == 0 and x in foundBoxes:
openBoxes.append(x)
status[x] = 1
return sum([candies[i] for i in openBoxes]) | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR NUMBER FOR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
N = len(status)
queue = list()
unopened = set()
for i in initialBoxes:
if status[i] == 1:
queue.append(i)
else:
unopened.add(i)
candiesCount = 0
while len(queue) > 0:
top = queue.pop(0)
candiesCount += candies[top]
for k in keys[top]:
status[k] = 1
if {k}.issubset(unopened):
queue.append(k)
unopened.remove(k)
for c in containedBoxes[top]:
if status[c] == 1:
queue.append(c)
else:
unopened.add(c)
return candiesCount | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
if len(initialBoxes) == 0:
return 0
else:
boxdict = {}
candysum = 0
for initialbox in initialBoxes:
boxdict[initialbox] = [0, 0]
p = len(status) ** 2
while p > 0:
for i in range(len(containedBoxes)):
if i in boxdict:
for containedbox in containedBoxes[i]:
boxdict[containedbox] = [0, 0]
p -= 1
for j in range(len(keys)):
for boxedkeys in keys[j]:
if boxedkeys in boxdict:
boxdict[boxedkeys][1] = 1
for l in range(len(status)):
if status[l] == 1 and l in boxdict:
boxdict[l][0] = 1
for key in boxdict:
if (
boxdict[key][0] == 1
and boxdict[key][1] == 1
or (boxdict[key][0] == 1 or boxdict[key][1] == 1)
):
candysum += candies[key]
return candysum | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR LIST NUMBER NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER WHILE VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR LIST NUMBER NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR IF VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR VAR NUMBER NUMBER VAR VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
n = len(status)
hold = initialBoxes
res = 0
key = set()
flag = True
while flag:
nxt = []
flag = False
for i in range(len(hold)):
cur = hold.pop()
if status[cur] == 1 or status[cur] == 0 and cur in key:
flag = True
res += candies[cur]
for k in keys[cur]:
key.add(k)
for b in containedBoxes[cur]:
nxt.append(b)
else:
nxt.append(cur)
hold = nxt
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR NUMBER VAR VAR NUMBER VAR VAR ASSIGN VAR NUMBER VAR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
locked_boxes = set()
unused_keys = set()
candy_count = 0
cur_box = []
for box in initialBoxes:
if status[box] or box in unused_keys:
candy_count += candies[box]
for i in keys[box]:
if not status[i]:
if i not in locked_boxes:
unused_keys.add(i)
else:
locked_boxes.remove(i)
cur_box.append(i)
for i in containedBoxes[box]:
if not status[i] and i not in unused_keys:
locked_boxes.add(i)
else:
cur_box.append(i)
else:
locked_boxes.add(box)
while cur_box:
box = cur_box.pop()
candy_count += candies[box]
for i in keys[box]:
if not status[i]:
if i not in locked_boxes:
unused_keys.add(i)
else:
locked_boxes.remove(i)
cur_box.append(i)
for i in containedBoxes[box]:
if not status[i] and i not in unused_keys:
locked_boxes.add(i)
else:
cur_box.append(i)
if i in unused_keys:
unused_keys.remove(i)
if box in unused_keys:
unused_keys.remove(box)
return candy_count | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR VAR VAR VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
active_set = initialBoxes
total_candies = 0
changed = True
while changed:
changed = False
next_active_set = []
for b in active_set:
if status[b]:
changed = True
total_candies += candies[b]
for k in keys[b]:
status[k] = 1
next_active_set += containedBoxes[b]
else:
next_active_set.append(b)
active_set = next_active_set
return total_candies
def SLOWmaxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
open_boxes = set()
closed_boxes = set()
for b in initialBoxes:
if status[b]:
open_boxes.add(b)
else:
closed_boxes.add(b)
total_candies = 0
while len(open_boxes) > 0:
b = open_boxes.pop()
total_candies += candies[b]
for key in keys[b]:
status[key] = 1
if key in closed_boxes:
closed_boxes.remove(key)
open_boxes.add(key)
for bb in containedBoxes[b]:
if status[bb]:
open_boxes.add(bb)
else:
closed_boxes.add(bb)
return total_candies | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR RETURN VAR VAR FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(self, status, candies, keys, containedBoxes, initialBoxes):
n = len(status)
access = [0] * n
source = []
visited = set()
for i in initialBoxes:
access[i] = 1
visited.add(i)
source.append(i)
ans = 0
while source:
i = source.pop()
ans += candies[i]
for j in containedBoxes[i]:
access[j] = 1
for j in keys[i]:
status[j] = 1
for j in containedBoxes[i]:
if j not in visited and status[j]:
visited.add(j)
source.append(j)
for j in keys[i]:
if j not in visited and access[j]:
visited.add(j)
source.append(j)
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
q = collections.deque()
keys_set = set([i for i in range(len(status)) if status[i] == 1])
boxes_set = set(initialBoxes[:])
for box in initialBoxes:
if box in keys_set:
q.append(box)
res = 0
visited = set()
while q:
boxi = q.popleft()
if boxi in visited:
continue
else:
visited.add(boxi)
res += candies[boxi]
for nei in containedBoxes[boxi]:
if nei in keys_set:
q.append(nei)
boxes_set.add(nei)
for key in keys[boxi]:
if key in boxes_set and key not in keys_set:
q.append(key)
keys_set.add(key)
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
n = len(status)
self.count = 0
queue = collections.deque([i for i in initialBoxes if status[i] == 1])
boxSet = set(initialBoxes)
keySet = set(i for i in range(n) if status[i] == 1)
opened = set()
while queue:
l = len(queue)
for i in range(l):
cur = queue.popleft()
self.count += candies[cur]
opened.add(cur)
for key in keys[cur]:
keySet.add(key)
if key not in opened and key in boxSet and key not in queue:
queue.append(key)
for box in containedBoxes[cur]:
boxSet.add(box)
if box not in opened and box in keySet and box not in queue:
queue.append(box)
return self.count | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
totalBox = len(status)
gotBoxes = set(initialBoxes)
i = 0
ans = 0
visited = set()
while i < len(initialBoxes):
boxNow = initialBoxes[i]
if boxNow not in visited:
if status[boxNow] == 1:
visited.add(boxNow)
ans += candies[boxNow]
for newBox in containedBoxes[boxNow]:
if status[newBox] == 1:
initialBoxes.append(newBox)
else:
gotBoxes.add(newBox)
for keyBox in keys[boxNow]:
status[keyBox] = 1
if keyBox in gotBoxes:
initialBoxes.append(keyBox)
gotBoxes.remove(keyBox)
else:
gotBoxes.add(boxNow)
i += 1
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
total = 0
boxes = initialBoxes[:]
locked = []
while boxes:
if not any([status[temp] for temp in boxes]):
break
box = boxes.pop(0)
if status[box]:
total += candies[box]
for contained in containedBoxes[box]:
boxes.append(contained)
for key in keys[box]:
status[key] = 1
else:
boxes.append(box)
return total | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR LIST WHILE VAR IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
q = collections.deque()
keys_set = set()
boxes_set = set()
for box in initialBoxes:
if status[box] == 1 or box in keys_set:
q.append(box)
for key in keys[box]:
keys_set.add(key)
if key in boxes_set:
q.append(key)
boxes_set.remove(key)
else:
boxes_set.add(box)
res = 0
visited = set()
while q:
boxi = q.popleft()
if boxi in visited:
continue
else:
visited.add(boxi)
res += candies[boxi]
for nei in containedBoxes[boxi]:
if status[nei] == 1 or nei in keys_set:
q.append(nei)
for key in keys[nei]:
keys_set.add(key)
if key in boxes_set:
q.append(key)
boxes_set.remove(key)
else:
boxes_set.add(nei)
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
found = [0] * len(status)
hasKeys = status
queue = collections.deque()
for box in initialBoxes:
found[box] = 1
if hasKeys[box]:
queue.append(box)
res = 0
while queue:
box = queue.popleft()
res += candies[box]
for t in containedBoxes[box]:
found[t] = 1
if hasKeys[t]:
queue.append(t)
for t in keys[box]:
if not hasKeys[t] and found[t]:
queue.append(t)
hasKeys[t] = 1
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
q = collections.deque()
for box in initialBoxes:
q.append(box)
keys_found = set()
ans = 0
found = True
while q and found:
size = len(q)
found = False
for _ in range(size):
cur = q.popleft()
if status[cur] == 0 and cur not in keys_found:
q.append(cur)
else:
found = True
ans += candies[cur]
for box in containedBoxes[cur]:
q.append(box)
for key in keys[cur]:
keys_found.add(key)
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(self, status, candies, keys, containedBoxes, initialBoxes):
ret = 0
collected_keys = set()
opened = set()
queue = initialBoxes
while queue:
cont = False
for i in range(len(queue)):
box_index = queue.pop(0)
if status[box_index] or box_index in collected_keys:
if box_index not in opened:
cont = True
opened.add(box_index)
ret += candies[box_index]
queue += containedBoxes[box_index]
collected_keys.update(keys[box_index])
else:
queue.append(box_index)
if not cont:
break
return ret | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR WHILE VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR VAR IF VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR RETURN VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
if not initialBoxes:
return 0
if 1 not in status:
return 0
boxes = []
result = 0
while True:
flag = False
for b in initialBoxes:
if status[b] == 1:
flag = True
result += candies[b]
boxes += containedBoxes[b]
for j in keys[b]:
status[j] = 1
initialBoxes.remove(b)
initialBoxes += boxes
boxes = []
if not flag:
break
return result | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR RETURN NUMBER IF NUMBER VAR RETURN NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR LIST IF VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
if not initialBoxes:
return 0
unused_keys = set()
boxes = set([i for i in initialBoxes if status[i] == 0])
Q = collections.deque([i for i in initialBoxes if status[i] == 1])
res = 0
while Q:
cur = Q.popleft()
res += candies[cur]
for k in keys[cur]:
if k in boxes:
Q.append(k)
boxes.discard(k)
else:
unused_keys.add(k)
for b in containedBoxes[cur]:
if b in unused_keys or status[b] == 1:
Q.append(b)
unused_keys.discard(b)
else:
boxes.add(b)
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
open = initialBoxes
x_keys = set()
locked = set()
result = 0
seen = set()
while len(open) != 0:
i = open.pop(0)
if i in seen:
continue
for key in keys[i]:
x_keys.add(key)
for box in containedBoxes[i]:
if box in x_keys or status[box] == 1:
open.append(box)
else:
locked.add(box)
for box in locked:
if box in x_keys:
open.append(box)
result += candies[i]
seen.add(i)
return result | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
n = len(status)
boxes = []
unboxes = set()
for b in initialBoxes:
if status[b]:
boxes.append(b)
else:
unboxes.add(b)
visited = set()
got = 0
while boxes:
ns = set()
for box in boxes:
if box not in visited:
got += candies[box]
visited.add(box)
for key in keys[box]:
status[key] |= 1
for box in boxes:
for adj in containedBoxes[box]:
unboxes.add(adj)
newBoxes = set()
for box in unboxes:
if status[box] and box not in visited:
ns.add(box)
newBoxes.add(box)
for box in newBoxes:
unboxes.remove(box)
boxes = list(ns)
return got | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR VAR VAR NUMBER FOR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
def dfs(i):
if i in visited:
return
if status[i] == 0:
missedKey.add(i)
return
visited.add(i)
self.ans += candies[i]
for k in keys[i]:
status[k] = 1
if k in missedKey:
missedKey.discard(k)
dfs(k)
for j in containedBoxes[i]:
dfs(j)
visited = set()
missedKey = set()
self.ans = 0
for i in initialBoxes:
dfs(i)
return self.ans | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FUNC_DEF IF VAR VAR RETURN IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN EXPR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
done = set()
to_open = set()
current_keys = set()
opened = list()
for box in initialBoxes:
if status[box] == 1:
opened.append(box)
else:
to_open.add(box)
total = 0
while opened:
box = opened.pop()
if box in done:
continue
total += candies[box]
done.add(box)
for openable in keys[box]:
if openable in to_open:
opened.append(openable)
to_open.remove(openable)
else:
current_keys.add(openable)
for contained in containedBoxes[box]:
if status[contained] == 1:
opened.append(contained)
elif contained in current_keys:
opened.append(contained)
current_keys.remove(contained)
else:
to_open.add(contained)
return total | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
maximum = 0
cur = []
closeddic = dict()
size = 0
for i in initialBoxes:
if status[i] == 1:
cur.append(i)
size += 1
else:
closeddic[i] = 1
consumed = dict()
keysdic = dict()
while size > 0:
nex = []
nsize = 0
for i in cur:
maximum += candies[i]
consumed[i] = True
for j in containedBoxes[i]:
if (status[j] == 1 or j in keysdic) and j not in consumed:
nex.append(j)
nsize += 1
elif status[j] == 0:
closeddic[j] = 1
for j in keys[i]:
keysdic[j] = 1
if j in closeddic and j not in consumed:
del closeddic[j]
nex.append(j)
nsize += 1
size = nsize
cur = nex
return maximum | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
self.keys_found = set()
self.boxes_found = set()
self.boxes_found.update(initialBoxes)
n = len(status)
visited = [0] * n
for box in initialBoxes:
self.dfs(box, keys, containedBoxes, visited)
ans = 0
for i in range(n):
if i in self.boxes_found and (i in self.keys_found or status[i] == 1):
ans += candies[i]
return ans
def dfs(self, cur, keys, containedBoxes, visited):
if visited[cur] == 1:
return
visited[cur] = 1
boxes = containedBoxes[cur]
self.keys_found.update(keys[cur])
self.boxes_found.update(boxes)
for box in boxes:
self.dfs(box, keys, containedBoxes, visited) | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR NUMBER VAR VAR VAR RETURN VAR VAR FUNC_DEF IF VAR VAR NUMBER RETURN ASSIGN VAR VAR NUMBER ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
q = []
res = 0
vis = set()
collected = set()
for i in initialBoxes:
q.append(i)
while q:
cur = q.pop(0)
vis.add(cur)
if status[cur] != 1 or cur in collected:
continue
res += candies[cur]
collected.add(cur)
for i in keys[cur]:
status[i] = 1
if i in vis:
q.append(i)
for j in containedBoxes[cur]:
q.append(j)
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR WHILE VAR ASSIGN VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
stack = deque(initialBoxes)
seen = set()
kseen = set()
while stack:
for _ in range(len(stack)):
box = stack.popleft()
seen.add(box)
for key in keys[box]:
status[key] = 1
for tbox in containedBoxes[box]:
if tbox not in seen:
stack.append(tbox)
res = 0
for box in seen:
if candies[box] and status[box]:
res += candies[box]
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR WHILE VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR VAR VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(self, status, candies, keys, containedBoxes, initialBoxes):
boxes = set(initialBoxes)
bfs = [i for i in boxes if status[i]]
for i in bfs:
for j in containedBoxes[i]:
boxes.add(j)
if status[j]:
bfs.append(j)
for j in keys[i]:
if status[j] == 0 and j in boxes:
bfs.append(j)
status[j] = 1
return sum(candies[i] for i in bfs) | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR FOR VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
keyHold = set([idx for idx, b in enumerate(status) if b == 1])
boxHold = set(initialBoxes)
opened = set()
while True:
flag = False
newKey = set()
newBox = set()
for b in boxHold:
if b in keyHold:
opened.add(b)
flag = True
for k in keys[b]:
newKey.add(k)
for nb in containedBoxes[b]:
if nb not in opened:
newBox.add(nb)
if not flag:
break
for k in newKey:
keyHold.add(k)
for b in newBox:
boxHold.add(b)
for b in opened:
boxHold.discard(b)
return sum([c for idx, c in enumerate(candies) if idx in opened]) | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR WHILE NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR IF VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
d = dict()
arr = []
key = []
for i in range(0, len(status)):
arr.append(-1)
key.append(-1)
d[i] = [status[i], candies[i], keys[i], containedBoxes[i]]
if status[i] == 1:
key[-1] = 1
c = 0
res = []
while 1 != -1:
tmp = []
for x in initialBoxes:
if arr[x] == -1 and key[x] == 1:
arr[x] = 1
c += d[x][1]
for y in d[x][2]:
key[y] = 1
for y in d[x][3]:
if key[y] == 1:
tmp.append(y)
else:
res.append(y)
elif key[x] == -1:
res.append(x)
initialBoxes = []
for y in res:
if key[y] == 1 and arr[y] == -1:
initialBoxes.append(y)
initialBoxes += tmp
if initialBoxes == []:
break
return c | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR LIST VAR VAR VAR VAR VAR VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST WHILE NUMBER NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR VAR VAR NUMBER FOR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR IF VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR IF VAR LIST RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
maxCandy = 0
keyset = set()
numbox = len(status)
q = list(initialBoxes)
unopenedBOXES = []
while q:
for _ in range(len(q)):
thisbox = q.pop(0)
if status[thisbox] == 1 or thisbox in keyset:
maxCandy += candies[thisbox]
candies[thisbox] = 0
for key in keys[thisbox]:
keyset.add(key)
else:
unopenedBOXES.append(thisbox)
for x in containedBoxes[thisbox]:
q.append(x)
for unopenedBOX in unopenedBOXES:
if unopenedBOX in keyset:
maxCandy += candies[unopenedBOX]
return maxCandy | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST WHILE VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR VAR NUMBER VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR IF VAR VAR VAR VAR VAR RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
parent = collections.defaultdict(int)
for i, row in enumerate(containedBoxes):
for j in row:
parent[j] = i
n = len(status)
bfs = collections.deque([(i, False) for i in initialBoxes])
res = 0
seen = set()
while bfs:
box, st = bfs.popleft()
if box in seen:
continue
if status[box] == 0:
if st:
return res
bfs.append((box, True))
continue
seen.add(box)
res += candies[box]
for key in keys[box]:
status[key] = 1
for cbox in containedBoxes[box]:
bfs.append((cbox, False))
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR WHILE VAR ASSIGN VAR VAR FUNC_CALL VAR IF VAR VAR IF VAR VAR NUMBER IF VAR RETURN VAR EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER RETURN VAR VAR |
Given n boxes, each box is given in the format [status, candies, keys, containedBoxes] where:
status[i]: an integer which is 1 if box[i] is open and 0 if box[i] is closed.
candies[i]: an integer representing the number of candies in box[i].
keys[i]: an array contains the indices of the boxes you can open with the key in box[i].
containedBoxes[i]: an array contains the indices of the boxes found in box[i].
You will start with some boxes given in initialBoxes array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.
Return the maximum number of candies you can get following the rules above.
Example 1:
Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
Output: 16
Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.
In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.
Total number of candies collected = 7 + 4 + 5 = 16 candy.
Example 2:
Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]
Output: 6
Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.
Example 3:
Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
Output: 1
Example 4:
Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
Output: 0
Example 5:
Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
Output: 7
Constraints:
1 <= status.length <= 1000
status.length == candies.length == keys.length == containedBoxes.length == n
status[i] is 0 or 1.
1 <= candies[i] <= 1000
0 <= keys[i].length <= status.length
0 <= keys[i][j] < status.length
All values in keys[i] are unique.
0 <= containedBoxes[i].length <= status.length
0 <= containedBoxes[i][j] < status.length
All values in containedBoxes[i] are unique.
Each box is contained in one box at most.
0 <= initialBoxes.length <= status.length
0 <= initialBoxes[i] < status.length | class Solution:
def maxCandies(
self,
status: List[int],
candies: List[int],
keys: List[List[int]],
containedBoxes: List[List[int]],
initialBoxes: List[int],
) -> int:
q, res = collections.deque(initialBoxes), 0
while q:
l = len(q)
for _ in range(l):
b, changed = q.popleft(), False
if status[b]:
changed = True
res += candies[b]
q.extend(containedBoxes[b])
for i in keys[b]:
status[i] = 1
else:
q.append(b)
if not changed:
return res
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR RETURN VAR RETURN VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | def power(n):
if n == 0:
return 1
else:
tmp = power(n // 2) % 1000000007 % 1000000007
if n % 2 == 1:
return 26 * tmp * tmp % 1000000007
else:
return tmp * tmp % 1000000007
def powernext(n):
if n == 0:
return 1
else:
tmp = powernext(n // 2) % 1000000007 % 1000000007
if n % 2 == 1:
return 25 * tmp * tmp % 1000000007
else:
return tmp * tmp % 1000000007
a = powernext(1000000005) % 1000000007
t = int(input())
for i in range(t):
n = int(input())
if n % 2 == 1:
print((power(n // 2 + 1) + 52 * (power(n // 2) - 1) * a) % 1000000007)
else:
print(52 * (power(n // 2) - 1) * a % 1000000007) | FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP BIN_OP NUMBER VAR VAR NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER FUNC_DEF IF VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP BIN_OP NUMBER VAR VAR NUMBER RETURN BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER BIN_OP BIN_OP NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | M = 10**9 + 7
for _ in range(int(input())):
n = int(input())
ans = 52 * (pow(26, n // 2, M) - 1) * pow(25, M - 2, M)
if n % 2 == 1:
ans += pow(26, n // 2 + 1, M)
print(ans % M) | ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | mod = 1000000007
for _ in range(int(input())):
n = int(input())
if n == 1:
print(26)
elif n == 2:
print(52)
else:
k = -(-n // 2)
if n % 2 == 0:
ans = 26 * (
1 + 26 * (pow(26, k - 1, mod) - 1) * pow(26 - 1, mod - 2, mod)
) + 26 * (1 + 26 * (pow(26, k - 1, mod) - 1) * pow(26 - 1, mod - 2, mod))
else:
ans = 26 * (
1 + 26 * (pow(26, k - 1, mod) - 1) * pow(26 - 1, mod - 2, mod)
) + 26 * (1 + 26 * (pow(26, k - 2, mod) - 1) * pow(26 - 1, mod - 2, mod))
print(ans % mod) | ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP NUMBER BIN_OP BIN_OP NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP NUMBER NUMBER BIN_OP VAR NUMBER VAR BIN_OP NUMBER BIN_OP NUMBER BIN_OP BIN_OP NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP NUMBER NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP NUMBER BIN_OP BIN_OP NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP NUMBER NUMBER BIN_OP VAR NUMBER VAR BIN_OP NUMBER BIN_OP NUMBER BIN_OP BIN_OP NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP NUMBER NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | mdl = 1000000007
i25 = pow(25, mdl - 2, mdl)
for _ in range(int(input())):
n = int(input())
k = n // 2 + 1
ko = ke = pow(26, k, mdl) - 1
if 1 == n % 2:
ko = ((ko + 1) * 26 - 1) % mdl
tot = ((ko + ke) * i25 - 2) % mdl
if tot < 0:
tot += mdl
print(tot) | ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR BIN_OP FUNC_CALL VAR NUMBER VAR VAR NUMBER IF NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | def topowermod(v, e, m):
op = 1
v %= m
be = bin(e)
for c in be[2:]:
op = op * op % m
if "1" == c:
op = op * v % m
return op
def modinv(v, m):
v %= m
tb = [(m, 1, 0), (v, 0, 1)]
while tb[-1][0] > 0:
q = tb[-2][0] // tb[-1][0]
nr = list(tb[-2])
for e in range(3):
nr[e] -= q * tb[-1][e]
tb.append(tuple(nr))
if tb[-2][0] == 1:
a = tb[-2][2]
while a < 0:
a += m
return a % m
mdl = 1000000007
i25 = modinv(25, mdl)
for _ in range(int(input())):
n = int(input())
k = n // 2
ke = topowermod(26, k + 1, mdl)
ko = ke
if 1 == n % 2:
ko = 26 * ke % mdl
ans = ((ko - 1 + ke - 1) * i25 + mdl - 2) % mdl
print(ans) | FUNC_DEF ASSIGN VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF STRING VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR RETURN VAR FUNC_DEF VAR VAR ASSIGN VAR LIST VAR NUMBER NUMBER VAR NUMBER NUMBER WHILE VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR VAR BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER WHILE VAR NUMBER VAR VAR RETURN BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR VAR IF NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | T = int(input())
inv = 280000002
mod = 1000000007
for i in range(T):
n = int(input())
if n % 2 == 0:
n = n // 2 + 1
a = 2 * ((pow(26, n, mod) - 1) * inv - 1) % mod
print(a)
else:
n = n // 2 + 2
a = 2 * ((pow(26, n, mod) - 1) * inv - 1) % mod
a = (a - pow(26, n - 1, mod)) % mod
print(a) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP BIN_OP BIN_OP FUNC_CALL VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP BIN_OP BIN_OP FUNC_CALL VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | def exp(a, n):
if n == 1:
return a
elif n % 2:
temp = exp(a, n // 2)
return temp % 1000000007 * temp % 1000000007 * a % 1000000007
else:
temp = exp(a, n // 2)
return temp % 1000000007 * temp % 1000000007 % 1000000007
def ans(n):
if n == 1:
return 26
elif n == 2:
return 52
if n % 2 == 0:
temp = exp(26, n / 2)
return (temp - 1) * 560000006 % 1000000007
else:
temp = exp(26, (n + 1) / 2)
return ((temp - 1) * 560000006 - temp) % 1000000007
t = int(input())
for index in range(0, t):
n = int(input())
print(ans(n)) | FUNC_DEF IF VAR NUMBER RETURN VAR IF BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER NUMBER FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR NUMBER RETURN NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | mod = 10**9 + 7
inv = pow(25, mod - 2, mod)
for i in range(int(input())):
N = int(input())
if N % 2 == 0:
r = 2 * 26 * (pow(26, N // 2, mod) - 1) % mod * inv % mod
else:
r = (27 * pow(26, N // 2 + 1, mod) - 52) % mod * inv % mod
print(r) | ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP NUMBER FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | def permutation(n, p):
r = 26
if n == 1:
return 26
elif n == 2:
return 52
elif n == 3:
return 728
elif n % 2 == 0:
return 2 * (bin_expo(r, n // 2 + 1, p) - r) * bin_expo(25, 1000000005, p) % p
else:
n = n + 1
return (
2 * ((bin_expo(r, n // 2 + 1, p) - r) * bin_expo(r - 1, 1000000005, p))
- bin_expo(26, n // 2, p)
) % p
def bin_expo(x, n, p):
if n == 0:
return 1
elif n == 1:
return x % p
else:
temp = bin_expo(x, n // 2, p)
temp = temp * temp % p
if n % 2 == 0:
return temp
else:
return x % p * temp % p
test = int(input())
for _ in range(test):
n = int(input())
p = 1000000007
print(int(permutation(n, p))) | FUNC_DEF ASSIGN VAR NUMBER IF VAR NUMBER RETURN NUMBER IF VAR NUMBER RETURN NUMBER IF VAR NUMBER RETURN NUMBER IF BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP BIN_OP NUMBER BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR FUNC_CALL VAR NUMBER NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN BIN_OP BIN_OP BIN_OP NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR VAR FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR NUMBER RETURN BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER RETURN VAR RETURN BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | T = int(input())
for t in range(T):
N = int(input())
if N % 2 == 0:
print(int(560000006 * (pow(26, N // 2, 1000000007) - 1)) % 1000000007)
else:
print(
(
int(560000006 * (pow(26, N // 2, 1000000007) - 1))
+ pow(26, N // 2 + 1, 1000000007)
)
% 1000000007
) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR BIN_OP NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER NUMBER FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER NUMBER |
Do you know that The Chef has a special interest in palindromes? Yes he does! Almost all of the dishes in his restaurant is named by a palindrome strings. The problem is that a name of a dish should not be too long, so The Chef has only limited choices when naming a new dish.
For the given positive integer N, your task is to calculate the number of palindrome strings of length not exceeding N, that contain only lowercase letters of English alphabet (letters from 'a' to 'z', inclusive). Recall that a palindrome is a string that reads the same left to right as right to left (as in "radar").
For example:
- For N = 1, we have 26 different palindromes of length not exceeding N:
"a", "b", ..., "z".
- For N = 2 we have 52 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz".
- For N = 3 we have 728 different palindromes of length not exceeding N:
"a", "b", ..., "z",
"aa", "bb", ..., "zz",
"aaa", "aba", ..., "aza",
"bab", "bbb", ..., "bzb",
...,
"zaz", "zbz", ..., "zzz".
Since the answer can be quite large you should output it modulo 1000000007 (109 + 7). Yes, we know, most of you already hate this modulo, but there is nothing we can do with it :)
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows. The only line of each test case contains a single integer N.
-----Output-----
For each test case, output a single line containing the answer for the corresponding test case.
-----Constrains-----
- 1 ≤ T ≤ 1000
- 1 ≤ N ≤ 109
-----Example-----
Input:
5
1
2
3
4
100
Output:
26
52
728
1404
508533804
-----Explanation-----
The first three examples are explained in the problem statement above. | mbase = 10**9 + 7
inv_25 = 280000002
t = int(input())
for it in range(t):
n = int(input())
m = n // 2
pow26mp1 = pow(26, m + 1, mbase)
res = 2 * (pow26mp1 - 26) * inv_25
if n % 2 == 1:
res += pow26mp1
res %= mbase
print(res) | ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR BIN_OP BIN_OP NUMBER BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
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