description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
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You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
i = 1
j = 10**6
ans = 10**9
while i <= j:
mid = i + (j - i) // 2
arr = [0] * (n + 1)
curr = 0
moves = 0
for ii in range(n):
if ii - w >= 0:
curr -= arr[ii - w]
if a[ii] + curr <= mid:
arr[ii] = mid - a[ii] - curr
curr += arr[ii]
moves += arr[ii]
if moves <= k:
ans = mid
i = mid + 1
else:
j = mid - 1
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
diff = [0] * n
mn = a[0]
for i in range(1, n):
diff[i] = a[i] - a[i - 1]
mn = min(mn, a[i])
mx = mn + k
ans = 0
def isValid(k, w, tmp, a):
for i in range(n):
if a[i] < tmp:
if tmp - a[i] > k:
return False
st = i
inc = tmp - a[i]
k -= inc
if st + w >= n:
st = max(0, n - w)
for j in range(st, min(n, st + w)):
a[j] += inc
return True
while mn <= mx:
tmp = (mn + mx) // 2
if isValid(k, w, tmp, a[:]):
ans = tmp
mn = tmp + 1
else:
mx = tmp - 1
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR IF BIN_OP VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR RETURN NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def helper(self, n, k, w, a, req):
res, needed = 0, 0
additive = [(0) for _ in range(n)]
for i in range(n):
if i > 0:
additive[i] += additive[i - 1]
if a[i] + additive[i] < req:
needed = req - a[i] - additive[i]
res += needed
additive[i] += needed
if i < n - w:
additive[i + w] -= needed
return res <= k
def maximizeMinHeight(self, n, k, w, a):
left, right = 1, 10000000000
mid = left + (right - left) // 2
while left < mid:
if self.helper(n, k, w, a, mid):
left = mid
else:
right = mid
mid = left + (right - left) // 2
return mid | CLASS_DEF FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
def func(arr):
dibba = [0] * n
ad, nt = 0, k
for i in range(n):
ad += dibba[i]
v = a[i] + ad
if v < arr:
res = arr - v
dibba[i] += res
if i + w < n:
dibba[i + w] -= res
ad += res
nt -= res
if nt < 0:
return False
return True
L = min(a)
R = L + k + 1
while L < R:
m = (L + R) // 2
if func(m):
L = m + 1
else:
R = m
return L - 1 | CLASS_DEF FUNC_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR VAR NUMBER VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN BIN_OP VAR NUMBER |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
def check(lvl):
days = 0
fls = [0] * (n + 1)
diff = max(0, lvl - a[0])
fls[0] = diff
fls[w] -= diff
days = diff
for i in range(1, n):
fls[i] += fls[i - 1]
cur_ht = a[i] + fls[i]
diff = max(0, lvl - cur_ht)
days += diff
fls[i] += diff
if i + w < n:
fls[i + w] -= diff
return days <= k
l = 0
h = 1000000000
res = -1
while l <= h:
mid = (l + h) // 2
if check(mid):
l = mid + 1
res = mid
else:
h = mid - 1
return res | CLASS_DEF FUNC_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
def check(t, mid):
lst = arr[:]
for i in range(n):
if lst[i] < mid:
temp = mid - lst[i]
if temp > t:
return False
t -= temp
lst[i] = mid
for j in range(i + 1, min(n, i + w)):
lst[j] += temp
return True
lo, hi = min(arr), min(arr) + k
res = None
while lo <= hi:
mid = (lo + hi) // 2
if check(k, mid):
res = mid
lo = mid + 1
else:
hi = mid - 1
return res | CLASS_DEF FUNC_DEF FUNC_DEF ASSIGN VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR RETURN NUMBER VAR VAR ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR NONE WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
if n == 1:
return a[0] + k
minh, maxh = a[0], a[0]
da = [0]
for i in range(1, n):
minh = min(a[i], minh)
maxh = max(a[i], maxh)
da.append(a[i] - a[i - 1])
ans = minh
maxh += k
while minh <= maxh:
mid = (minh + maxh) // 2
da = [0]
for i in range(1, n):
da.append(a[i] - a[i - 1])
cumsum = a[0]
k_left = k
flag = True
for i in range(n):
cumsum += da[i]
if cumsum >= mid:
continue
if cumsum + k_left < mid:
flag = False
break
step = mid - cumsum
k_left -= step
cumsum += step
if i + w < n:
da[i + w] -= step
if flag:
minh = mid + 1
ans = max(ans, mid)
else:
maxh = mid - 1
return ans | CLASS_DEF FUNC_DEF IF VAR NUMBER RETURN BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR IF VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def possible(self, mnheight, k, w, a):
days = 0
prefsum = 0
dp = [0] * (n + 1)
for i in range(n):
prefsum += dp[i]
height = a[i] + prefsum
if height < mnheight:
add = mnheight - height
days += add
prefsum += add
dp[i] += add
dp[min(n, i + w)] -= add
return days <= k
def maximizeMinHeight(self, n, k, w, a):
l = min(a) + 1
h = max(a) + k
target = l - 1
while l <= h:
temp = (l + h) // 2
if self.possible(temp, k, w, a):
target = temp
l = temp + 1
else:
h = temp - 1
return target | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
l, h = min(a), 10**15
def solve(mid):
pre_sum = [0] * n
cnt = 0
for i in range(n):
if i > 0:
pre_sum[i] += pre_sum[i - 1]
if a[i] + pre_sum[i] < mid:
gg = mid - (a[i] + pre_sum[i])
cnt += gg
pre_sum[i] += gg
if i + w < n:
pre_sum[i + w] -= gg
return cnt <= k
while l < h:
mid = l + h + 1 >> 1
if solve(mid):
l = mid
else:
h = mid - 1
return l | CLASS_DEF FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP NUMBER NUMBER FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR RETURN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
def helper(n, k, w, a, mid):
for i in range(n):
if a[i] < mid:
temp = mid - a[i]
if temp > k:
return False
k -= temp
j = i
while j < n and j < i + w:
a[j] += temp
j += 1
return True
mini = min(a)
maxi = mini + k
ans = 0
while mini <= maxi:
mid = (mini + maxi) // 2
if helper(n, k, w, a[:], mid):
mini = mid + 1
ans = mid
else:
maxi = mid - 1
return ans | CLASS_DEF FUNC_DEF FUNC_DEF FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR RETURN NUMBER VAR VAR ASSIGN VAR VAR WHILE VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def check(self, arr, k, w, mid):
ar = list(arr)
for i in range(len(arr)):
if ar[i] < mid:
rem = mid - ar[i]
if rem > k:
return False
k -= rem
for j in range(i, min(i + w, len(ar))):
ar[j] += rem
return True
def bin_search(self, start, end, arr, k, w):
if start <= end:
mid = (end + start) // 2
if self.check(arr, k, w, mid):
return self.bin_search(mid + 1, end, arr, k, w)
else:
return self.bin_search(start, mid - 1, arr, k, w)
return start - 1
def maximizeMinHeight(self, n, k, w, a):
return self.bin_search(min(a), min(a) + k, a, k, w) | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR RETURN NUMBER VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN NUMBER FUNC_DEF IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR VAR RETURN BIN_OP VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
mn = min(a)
mx = mn + k + 1
def f(ht):
days = 0
curr_ht = a[0]
flower = [0] * (n + 1)
diff = max(0, ht - curr_ht)
flower[0] += diff
days += diff
flower[w] -= diff
for i in range(1, n):
flower[i] += flower[i - 1]
curr_ht = a[i] + flower[i]
diff = max(0, ht - curr_ht)
flower[i] += diff
days += diff
if i + w < n:
flower[i + w] -= diff
return days <= k
while mn < mx:
ht = (mn + mx) // 2
if not f(ht):
mx = ht
else:
mn = ht + 1
return mn - 1 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR RETURN VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN BIN_OP VAR NUMBER |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def height_is_achievable(self, k, w, a, mid):
prefix = [0] * len(a)
for i in range(len(a)):
if a[i] + prefix[i] >= mid:
continue
diff = mid - a[i] - prefix[i]
k -= diff
if k < 0:
return False
for index in range(i, i + w):
if index >= len(prefix):
break
prefix[index] += diff
return True
def maximizeMinHeight(self, n, k, w, a):
max_height_achievable = max(a) + k
min_height = min(a)
while min_height <= max_height_achievable:
mid = (max_height_achievable + min_height) // 2
if self.height_is_achievable(k, w, a, mid):
min_height = mid + 1
else:
max_height_achievable = mid - 1
return max_height_achievable | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR IF VAR NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN NUMBER FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def check(self, diff, k2, w, t2):
f = 0
for i in range(len(diff)):
diff[i] += f
f = diff[i]
if diff[i] >= t2:
continue
k2 -= t2 - diff[i]
if k2 < 0:
return 0
if i + w < len(diff):
diff[i + w] -= t2 - diff[i]
f = t2
return 1
def maximizeMinHeight(self, n, k, w, a):
dif = [0] * n
dif[0] = a[0]
mi = a[0]
for i in range(1, n):
dif[i] = a[i] - a[i - 1]
mi = min(a[i], mi)
ma = mi + k
ans = 0
while mi <= ma:
t = (mi + ma) // 2
if self.check(dif.copy(), k, w, t):
ans = t
mi = t + 1
else:
ma = t - 1
return ans | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR VAR IF VAR NUMBER RETURN NUMBER IF BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR RETURN NUMBER FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def isValid(self, n, l, k, w, m):
pre = [0] * (n + 1)
temp = 0
for i in range(0, n):
pre[i] += pre[i - 1]
val = pre[i] + l[i]
if val < m:
pre[i] += m - val
pre[min(n, i + w)] -= m - val
temp += m - val
return temp <= k
def maximizeMinHeight(self, n, k, w, a):
f = min(a)
r = max(a) + k
while f <= r:
mid = (f + r) // 2
if self.isValid(n, a, k, w, mid):
f = mid + 1
else:
r = mid - 1
return r | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, ac):
low = 1
high = 10**14
while low < high:
mid = (low + high) // 2
l = [0] * n
a = ac.copy()
prefix_sum = 0
prefix_add = [0] * (n + 1)
counter = 0
for i in range(n):
if i >= w:
prefix_sum -= prefix_add[i - w]
if a[i] + prefix_sum < mid:
to_be_added = mid - a[i] - prefix_sum
start = min(n, i + w) - w
prefix_add[start] += to_be_added
prefix_sum += to_be_added
counter += to_be_added
if counter > k:
high = mid
else:
low = mid + 1
return low - 1 | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN BIN_OP VAR NUMBER |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, arr):
def helper(mid):
lis = arr.copy()
cnt = 0
for i in range(n):
if lis[i] < mid:
cnt += mid - lis[i]
add = mid - lis[i]
for j in range(i, min(n, i + w)):
lis[j] += add
return cnt <= k
l, r = 1, 10**9
ans = min(arr)
while l <= r:
mid = (l + r) // 2
if helper(mid):
ans = mid
l = mid + 1
else:
r = mid - 1
return ans | CLASS_DEF FUNC_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR RETURN VAR VAR ASSIGN VAR VAR NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
high = 10**9
low = 0
ans = -1
while low <= high:
mid = (low + high) // 2
if self.isPossible(a, w, mid, k):
low = mid + 1
ans = max(mid, ans)
else:
high = mid - 1
return ans
def isPossible(self, arr, l, maxHeight, days):
water = [0] * len(arr)
for i in range(len(arr)):
if i > 0:
water[i] = water[i - 1]
curHei = water[i] + arr[i]
if i >= l:
curHei -= water[i - l]
if curHei < maxHeight:
water[i] += maxHeight - curHei
days -= maxHeight - curHei
if days < 0:
return False
return True | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER RETURN NUMBER RETURN NUMBER |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def isValid(self, arr, val, w, m):
n = len(arr)
difSum = [0] * n
pref = 0
for i in range(n):
pref -= difSum[i]
req = max(0, val - arr[i] - pref)
m -= req
pref += req
if i + w < n:
difSum[i + w] += req
if m < 0:
return False
return True
def maximizeMinHeight(self, n, k, w, a):
l = 0
r = 10**9 + 1
while l < r - 1:
mid = (l + r) // 2
if self.isValid(a, mid, w, k):
l = mid
else:
r = mid - 1
while r >= l:
if self.isValid(a, r, w, k):
return r
r -= 1 | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR IF BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR IF VAR NUMBER RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER WHILE VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR RETURN VAR VAR NUMBER |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
df = [0] * (n + w)
df[0] = a[0]
for j in range(1, n):
df[j] = a[j] - a[j - 1]
def valid(x):
dk = k
diff = df[:]
prev = 0
for l in range(n):
cur = prev + diff[l]
if cur < x:
e = x - cur
diff[l] += e
diff[l + w] -= e
dk -= e
cur = x
if dk < 0:
return False
prev = cur
return True
lo, hi = 0, 10**10
while lo < hi:
mid = (lo + hi) // 2
if valid(mid):
lo = mid + 1
else:
hi = mid
return lo - 1 | CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR IF VAR NUMBER RETURN NUMBER ASSIGN VAR VAR RETURN NUMBER ASSIGN VAR VAR NUMBER BIN_OP NUMBER NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR RETURN BIN_OP VAR NUMBER |
You have a garden with n flowers lined up in a row. The height of ith flower is a_{i} units. You will water them for k days. In one day you can water w continuous flowers (you can do this only once in a single day). Whenever you water a flower its height increases by 1 unit. You need to maximize the height of the smallest flower all the time.
Example 1:
Input:
n=6
k=2
w=3
a[]={2,2,2,2,1,1}
Output:
2
Explanation:
Water last three flowers for first day &
first three flowers for second day.The
new heights will be {3,3,3,3,2,2}
Example 2:
Input:
n=2
k=5
w=1
a[]={5,8}
Output:
9
Explanation:
For the first four days water the first flower then
water the last flower once.
Your Task:
You don't need to read input or print anything. Your task is to complete the function maximizeMinHeight() which takes the array a[], its size N, integer K, and an integer W as input parameters and returns the maximum height possible for the smallest flower.
Expected Time Complexity: O(NLogN)
Expected Space Complexity: O(N)
Constraints:
1 <= n<= 10^{5}
1<=w<=n
1<=k<=10^{5}
1 <= a[i] <= 10^{9} | class Solution:
def maximizeMinHeight(self, n, k, w, a):
def check(m, arr, c):
for i in range(len(arr)):
if m > arr[i]:
x = m - arr[i]
c -= x
if c < 0:
return False
for j in range(i, min(i + w, n)):
arr[j] += x
return True
l = min(a)
h = min(a) + k
ans = min(a)
while l <= h:
m = (l + h) // 2
if check(m, a[:], k):
ans = max(m, ans)
l = m + 1
else:
h = m - 1
return ans | CLASS_DEF FUNC_DEF FUNC_DEF FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR VAR IF VAR NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR |
You are given an array of integers $b_1, b_2, \ldots, b_n$.
An array $a_1, a_2, \ldots, a_n$ of integers is hybrid if for each $i$ ($1 \leq i \leq n$) at least one of these conditions is true:
$b_i = a_i$, or
$b_i = \sum_{j=1}^{i} a_j$.
Find the number of hybrid arrays $a_1, a_2, \ldots, a_n$. As the result can be very large, you should print the answer modulo $10^9 + 7$.
-----Input-----
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).
The second line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($-10^9 \le b_i \le 10^9$).
It is guaranteed that the sum of $n$ for all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer: the number of hybrid arrays $a_1, a_2, \ldots, a_n$ modulo $10^9 + 7$.
-----Examples-----
Input
4
3
1 -1 1
4
1 2 3 4
10
2 -1 1 -2 2 3 -5 0 2 -1
4
0 0 0 1
Output
3
8
223
1
-----Note-----
In the first test case, the hybrid arrays are $[1, -2, 1]$, $[1, -2, 2]$, $[1, -1, 1]$.
In the second test case, the hybrid arrays are $[1, 1, 1, 1]$, $[1, 1, 1, 4]$, $[1, 1, 3, -1]$, $[1, 1, 3, 4]$, $[1, 2, 0, 1]$, $[1, 2, 0, 4]$, $[1, 2, 3, -2]$, $[1, 2, 3, 4]$.
In the fourth test case, the only hybrid array is $[0, 0, 0, 1]$. | import sys
from sys import stdin
mod = 10**9 + 7
class SegTree:
def __init__(self, N, first):
self.NO = 2 ** (N - 1).bit_length()
self.First = first
self.data = [first] * (2 * self.NO)
def calc(self, l, r):
return (l + r) % mod
def update(self, ind, x):
ind += self.NO - 1
self.data[ind] = x
while ind >= 0:
ind = (ind - 1) // 2
self.data[ind] = self.calc(self.data[2 * ind + 1], self.data[2 * ind + 2])
def query(self, l, r):
L = l + self.NO
R = r + self.NO
s = self.First
while L < R:
if R & 1:
R -= 1
s = self.calc(s, self.data[R - 1])
if L & 1:
s = self.calc(s, self.data[L - 1])
L += 1
L >>= 1
R >>= 1
return s
def get(self, ind):
ind += self.NO - 1
return self.data[ind]
tt = int(stdin.readline())
for loop in range(tt):
n = int(stdin.readline())
b = [0] + list(map(int, stdin.readline().split()))
c = [0]
for i in range(1, n + 1):
c.append(c[-1] + b[i])
ctoi = {}
tlis = []
for i in range(len(c)):
if c[i] not in ctoi:
ctoi[c[i]] = 0
tlis.append(c[i])
tlis.sort()
for i in range(len(tlis)):
ctoi[tlis[i]] = i
ST = SegTree(len(tlis), 0)
ST.update(ctoi[0], 1)
for i in range(1, n + 1):
ind = ctoi[c[i - 1]]
ST.update(ind, ST.query(0, len(tlis)))
print(ST.query(0, len(tlis))) | IMPORT ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER CLASS_DEF FUNC_DEF ASSIGN VAR BIN_OP NUMBER FUNC_CALL BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP LIST VAR BIN_OP NUMBER VAR FUNC_DEF RETURN BIN_OP BIN_OP VAR VAR VAR FUNC_DEF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR WHILE VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER VAR BIN_OP BIN_OP NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR WHILE VAR VAR IF BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER RETURN VAR FUNC_DEF VAR BIN_OP VAR NUMBER RETURN 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 BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR DICT ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR |
You are given an array of integers $b_1, b_2, \ldots, b_n$.
An array $a_1, a_2, \ldots, a_n$ of integers is hybrid if for each $i$ ($1 \leq i \leq n$) at least one of these conditions is true:
$b_i = a_i$, or
$b_i = \sum_{j=1}^{i} a_j$.
Find the number of hybrid arrays $a_1, a_2, \ldots, a_n$. As the result can be very large, you should print the answer modulo $10^9 + 7$.
-----Input-----
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).
The second line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($-10^9 \le b_i \le 10^9$).
It is guaranteed that the sum of $n$ for all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer: the number of hybrid arrays $a_1, a_2, \ldots, a_n$ modulo $10^9 + 7$.
-----Examples-----
Input
4
3
1 -1 1
4
1 2 3 4
10
2 -1 1 -2 2 3 -5 0 2 -1
4
0 0 0 1
Output
3
8
223
1
-----Note-----
In the first test case, the hybrid arrays are $[1, -2, 1]$, $[1, -2, 2]$, $[1, -1, 1]$.
In the second test case, the hybrid arrays are $[1, 1, 1, 1]$, $[1, 1, 1, 4]$, $[1, 1, 3, -1]$, $[1, 1, 3, 4]$, $[1, 2, 0, 1]$, $[1, 2, 0, 4]$, $[1, 2, 3, -2]$, $[1, 2, 3, 4]$.
In the fourth test case, the only hybrid array is $[0, 0, 0, 1]$. | for s in [*open(0)][2::2]:
C = [0]
D = {(0): 1}
S = 1
for n in map(int, s.split()):
C += (C[-1] + n,)
for n in C[1:-1]:
D[n], S = S, (2 * S - D.get(n, 0)) % (10**9 + 7)
print(S) | FOR VAR LIST FUNC_CALL VAR NUMBER NUMBER NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR DICT NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR FOR VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array of integers $b_1, b_2, \ldots, b_n$.
An array $a_1, a_2, \ldots, a_n$ of integers is hybrid if for each $i$ ($1 \leq i \leq n$) at least one of these conditions is true:
$b_i = a_i$, or
$b_i = \sum_{j=1}^{i} a_j$.
Find the number of hybrid arrays $a_1, a_2, \ldots, a_n$. As the result can be very large, you should print the answer modulo $10^9 + 7$.
-----Input-----
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).
The second line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($-10^9 \le b_i \le 10^9$).
It is guaranteed that the sum of $n$ for all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer: the number of hybrid arrays $a_1, a_2, \ldots, a_n$ modulo $10^9 + 7$.
-----Examples-----
Input
4
3
1 -1 1
4
1 2 3 4
10
2 -1 1 -2 2 3 -5 0 2 -1
4
0 0 0 1
Output
3
8
223
1
-----Note-----
In the first test case, the hybrid arrays are $[1, -2, 1]$, $[1, -2, 2]$, $[1, -1, 1]$.
In the second test case, the hybrid arrays are $[1, 1, 1, 1]$, $[1, 1, 1, 4]$, $[1, 1, 3, -1]$, $[1, 1, 3, 4]$, $[1, 2, 0, 1]$, $[1, 2, 0, 4]$, $[1, 2, 3, -2]$, $[1, 2, 3, 4]$.
In the fourth test case, the only hybrid array is $[0, 0, 0, 1]$. | T = int(input())
for t in range(T):
n = int(input())
bb = [int(x) for x in input().split()]
sums_bef_offset = {bb[0]: 1}
offset = 0
all_sums = 1
base = 1000000007
result = 0
for i in range(1, n):
b = bb[i]
sums_0 = sums_bef_offset.get(-offset, 0)
offset += b
prev_all_sums = all_sums
sums_bef_offset[b - offset] = prev_all_sums
all_sums = (2 * prev_all_sums - sums_0) % base
print(all_sums) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP NUMBER VAR VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array of integers $b_1, b_2, \ldots, b_n$.
An array $a_1, a_2, \ldots, a_n$ of integers is hybrid if for each $i$ ($1 \leq i \leq n$) at least one of these conditions is true:
$b_i = a_i$, or
$b_i = \sum_{j=1}^{i} a_j$.
Find the number of hybrid arrays $a_1, a_2, \ldots, a_n$. As the result can be very large, you should print the answer modulo $10^9 + 7$.
-----Input-----
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).
The second line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($-10^9 \le b_i \le 10^9$).
It is guaranteed that the sum of $n$ for all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer: the number of hybrid arrays $a_1, a_2, \ldots, a_n$ modulo $10^9 + 7$.
-----Examples-----
Input
4
3
1 -1 1
4
1 2 3 4
10
2 -1 1 -2 2 3 -5 0 2 -1
4
0 0 0 1
Output
3
8
223
1
-----Note-----
In the first test case, the hybrid arrays are $[1, -2, 1]$, $[1, -2, 2]$, $[1, -1, 1]$.
In the second test case, the hybrid arrays are $[1, 1, 1, 1]$, $[1, 1, 1, 4]$, $[1, 1, 3, -1]$, $[1, 1, 3, 4]$, $[1, 2, 0, 1]$, $[1, 2, 0, 4]$, $[1, 2, 3, -2]$, $[1, 2, 3, 4]$.
In the fourth test case, the only hybrid array is $[0, 0, 0, 1]$. | p = lambda: list(map(int, input().split()))
for t in range(p()[0]):
N = p()[0]
B = p()
C = [0] * (N + 1)
for i in range(N):
C[i + 1] = C[i] + B[i]
S = 1
D = dict()
D[0] = 1
for i in range(1, N):
D[C[i]], S = S, (2 * S - D.get(C[i], 0)) % (10**9 + 7)
print(S) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR VAR VAR BIN_OP BIN_OP BIN_OP NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER BIN_OP BIN_OP NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR |
You are given an array of integers $b_1, b_2, \ldots, b_n$.
An array $a_1, a_2, \ldots, a_n$ of integers is hybrid if for each $i$ ($1 \leq i \leq n$) at least one of these conditions is true:
$b_i = a_i$, or
$b_i = \sum_{j=1}^{i} a_j$.
Find the number of hybrid arrays $a_1, a_2, \ldots, a_n$. As the result can be very large, you should print the answer modulo $10^9 + 7$.
-----Input-----
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).
The second line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($-10^9 \le b_i \le 10^9$).
It is guaranteed that the sum of $n$ for all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer: the number of hybrid arrays $a_1, a_2, \ldots, a_n$ modulo $10^9 + 7$.
-----Examples-----
Input
4
3
1 -1 1
4
1 2 3 4
10
2 -1 1 -2 2 3 -5 0 2 -1
4
0 0 0 1
Output
3
8
223
1
-----Note-----
In the first test case, the hybrid arrays are $[1, -2, 1]$, $[1, -2, 2]$, $[1, -1, 1]$.
In the second test case, the hybrid arrays are $[1, 1, 1, 1]$, $[1, 1, 1, 4]$, $[1, 1, 3, -1]$, $[1, 1, 3, 4]$, $[1, 2, 0, 1]$, $[1, 2, 0, 4]$, $[1, 2, 3, -2]$, $[1, 2, 3, 4]$.
In the fourth test case, the only hybrid array is $[0, 0, 0, 1]$. | import sys
mod = 10**9 + 7
for t in range(int(input())):
N = int(input())
B = list(map(int, input().split()))
C = [0] * (N + 1)
for i in range(N):
C[i + 1] = C[i] + B[i]
S = 1
D = dict()
D[0] = 1
for i in range(N - 1):
T = (S + S - D.get(C[i + 1], 0)) % mod
D[C[i + 1]] = S
S = T
print(S) | IMPORT 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 FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR |
You are given an array of integers $b_1, b_2, \ldots, b_n$.
An array $a_1, a_2, \ldots, a_n$ of integers is hybrid if for each $i$ ($1 \leq i \leq n$) at least one of these conditions is true:
$b_i = a_i$, or
$b_i = \sum_{j=1}^{i} a_j$.
Find the number of hybrid arrays $a_1, a_2, \ldots, a_n$. As the result can be very large, you should print the answer modulo $10^9 + 7$.
-----Input-----
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$).
The second line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($-10^9 \le b_i \le 10^9$).
It is guaranteed that the sum of $n$ for all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer: the number of hybrid arrays $a_1, a_2, \ldots, a_n$ modulo $10^9 + 7$.
-----Examples-----
Input
4
3
1 -1 1
4
1 2 3 4
10
2 -1 1 -2 2 3 -5 0 2 -1
4
0 0 0 1
Output
3
8
223
1
-----Note-----
In the first test case, the hybrid arrays are $[1, -2, 1]$, $[1, -2, 2]$, $[1, -1, 1]$.
In the second test case, the hybrid arrays are $[1, 1, 1, 1]$, $[1, 1, 1, 4]$, $[1, 1, 3, -1]$, $[1, 1, 3, 4]$, $[1, 2, 0, 1]$, $[1, 2, 0, 4]$, $[1, 2, 3, -2]$, $[1, 2, 3, 4]$.
In the fourth test case, the only hybrid array is $[0, 0, 0, 1]$. | import sys
from sys import stdin
mod = 10**9 + 7
tt = int(stdin.readline())
for loop in range(tt):
n = int(stdin.readline())
b = [0] + list(map(int, stdin.readline().split()))
c = [0]
for i in range(1, n + 1):
c.append(c[-1] + b[i])
dic = {}
dic[0] = 1
ds = 1
for i in range(1, n + 1):
nc = c[i - 1]
if nc not in dic:
last = 0
else:
last = dic[nc]
dic[nc] = ds
ds -= last
ds += dic[nc]
ds %= mod
print(ds % mod) | IMPORT ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER 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 LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a_1, a_2, ..., a_{n} in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).
procedure bubbleSortGraph()
build a graph G with n vertices and 0 edges
repeat
swapped = false
for i = 1 to n - 1 inclusive do:
if a[i] > a[i + 1] then
add an undirected edge in G between a[i] and a[i + 1]
swap( a[i], a[i + 1] )
swapped = true
end if
end for
until not swapped
/* repeat the algorithm as long as swapped value is true. */
end procedure
For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.
-----Input-----
The first line of the input contains an integer n (2 ≤ n ≤ 10^5). The next line contains n distinct integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ n).
-----Output-----
Output a single integer — the answer to the problem.
-----Examples-----
Input
3
3 1 2
Output
2
-----Note-----
Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2]. | def CeilIndex(A, l, r, key):
while r - l > 1:
m = l + (r - l) // 2
if A[m] >= key:
r = m
else:
l = m
return r
def LIS(A, size):
tailTable = [(0) for i in range(size + 1)]
len = 0
tailTable[0] = A[0]
len = 1
for i in range(1, size):
if A[i] < tailTable[0]:
tailTable[0] = A[i]
elif A[i] > tailTable[len - 1]:
tailTable[len] = A[i]
len += 1
else:
tailTable[CeilIndex(tailTable, -1, len - 1, A[i])] = A[i]
return len
def main():
n = int(input())
arr = list(map(int, input().split()))
print(LIS(arr, n))
main() | FUNC_DEF WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR |
Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a_1, a_2, ..., a_{n} in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).
procedure bubbleSortGraph()
build a graph G with n vertices and 0 edges
repeat
swapped = false
for i = 1 to n - 1 inclusive do:
if a[i] > a[i + 1] then
add an undirected edge in G between a[i] and a[i + 1]
swap( a[i], a[i + 1] )
swapped = true
end if
end for
until not swapped
/* repeat the algorithm as long as swapped value is true. */
end procedure
For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.
-----Input-----
The first line of the input contains an integer n (2 ≤ n ≤ 10^5). The next line contains n distinct integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ n).
-----Output-----
Output a single integer — the answer to the problem.
-----Examples-----
Input
3
3 1 2
Output
2
-----Note-----
Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2]. | import sys
sys.setrecursionlimit(1000000)
def solve():
(n,) = rv()
(a,) = rl(1)
mem = [10000000] * (n + 1)
mem[0] = a[0]
for i in range(1, n):
left, right = 0, n - 1
while left < right:
mid = (left + right) // 2
if a[i] < mem[mid]:
right = mid
else:
left = mid + 1
mem[left] = a[i]
res = 0
for i in range(1, n):
if mem[i] != 10000000:
res = i
print(res + 1)
def rv():
return list(map(int, input().split()))
def rl(n):
return [list(map(int, input().split())) for _ in range(n)]
if sys.hexversion == 50594544:
sys.stdin = open("test.txt")
solve() | IMPORT EXPR FUNC_CALL VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR |
Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a_1, a_2, ..., a_{n} in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).
procedure bubbleSortGraph()
build a graph G with n vertices and 0 edges
repeat
swapped = false
for i = 1 to n - 1 inclusive do:
if a[i] > a[i + 1] then
add an undirected edge in G between a[i] and a[i + 1]
swap( a[i], a[i + 1] )
swapped = true
end if
end for
until not swapped
/* repeat the algorithm as long as swapped value is true. */
end procedure
For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.
-----Input-----
The first line of the input contains an integer n (2 ≤ n ≤ 10^5). The next line contains n distinct integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ n).
-----Output-----
Output a single integer — the answer to the problem.
-----Examples-----
Input
3
3 1 2
Output
2
-----Note-----
Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2]. | n = int(input())
a = list(map(int, input().split()))
bit = [0] * (n + 1)
def update(idx, val):
while idx <= n:
bit[idx] = max(bit[idx], val)
idx += idx & -idx
def query(idx):
res = 0
while idx > 0:
res = max(res, bit[idx])
idx -= idx & -idx
return res
b = []
for i in range(n):
b.append([a[i], -(i + 1)])
b.sort()
for i in range(n):
qe = query(-b[i][1])
update(-b[i][1], qe + 1)
print(query(n)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF WHILE VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR LIST VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a_1, a_2, ..., a_{n} in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).
procedure bubbleSortGraph()
build a graph G with n vertices and 0 edges
repeat
swapped = false
for i = 1 to n - 1 inclusive do:
if a[i] > a[i + 1] then
add an undirected edge in G between a[i] and a[i + 1]
swap( a[i], a[i + 1] )
swapped = true
end if
end for
until not swapped
/* repeat the algorithm as long as swapped value is true. */
end procedure
For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.
-----Input-----
The first line of the input contains an integer n (2 ≤ n ≤ 10^5). The next line contains n distinct integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ n).
-----Output-----
Output a single integer — the answer to the problem.
-----Examples-----
Input
3
3 1 2
Output
2
-----Note-----
Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2]. | MXusl = int
MXuso = input
MXusL = map
MXusr = min
MXusI = print
n = MXusl(MXuso())
a = [1000000.0] * (n + 1)
s = 1
for x in MXusL(MXusl, MXuso().split()):
l = 0
r = s
while r - l > 1:
m = l + r >> 1
if a[m] < x:
l = m
else:
r = m
s += r == s
a[r] = MXusr(a[r], x)
MXusI(s - 1) | ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER |
Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a_1, a_2, ..., a_{n} in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).
procedure bubbleSortGraph()
build a graph G with n vertices and 0 edges
repeat
swapped = false
for i = 1 to n - 1 inclusive do:
if a[i] > a[i + 1] then
add an undirected edge in G between a[i] and a[i + 1]
swap( a[i], a[i + 1] )
swapped = true
end if
end for
until not swapped
/* repeat the algorithm as long as swapped value is true. */
end procedure
For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.
-----Input-----
The first line of the input contains an integer n (2 ≤ n ≤ 10^5). The next line contains n distinct integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ n).
-----Output-----
Output a single integer — the answer to the problem.
-----Examples-----
Input
3
3 1 2
Output
2
-----Note-----
Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2]. | from sys import stdin
input = stdin.readline
class BIT:
def __init__(self, n):
self.n = n
self.tree = [0] * (n + 1)
def sum(self, i):
ans = 0
i += 1
while i > 0:
ans = max(ans, self.tree[i])
i -= i & -i
return ans
def update(self, i, value):
i += 1
while i <= self.n:
self.tree[i] = max(value, self.tree[i])
i += i & -i
def f(a):
newind = 0
maxs = 0
ans = 0
ft = BIT(2 * 10**5 + 5)
for i in a:
maxs = ft.sum(i - 1)
ft.update(i, maxs + 1)
ans = max(ans, maxs + 1)
return ans
a = input()
l = list(map(int, input().strip().split()))
print(f(l)) | ASSIGN VAR VAR CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER VAR NUMBER WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF VAR NUMBER WHILE VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP NUMBER BIN_OP NUMBER NUMBER NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a_1, a_2, ..., a_{n} in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).
procedure bubbleSortGraph()
build a graph G with n vertices and 0 edges
repeat
swapped = false
for i = 1 to n - 1 inclusive do:
if a[i] > a[i + 1] then
add an undirected edge in G between a[i] and a[i + 1]
swap( a[i], a[i + 1] )
swapped = true
end if
end for
until not swapped
/* repeat the algorithm as long as swapped value is true. */
end procedure
For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.
-----Input-----
The first line of the input contains an integer n (2 ≤ n ≤ 10^5). The next line contains n distinct integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ n).
-----Output-----
Output a single integer — the answer to the problem.
-----Examples-----
Input
3
3 1 2
Output
2
-----Note-----
Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2]. | def lis(a):
b = []
for c in a:
if len(b) == 0 or c > b[-1]:
b.append(c)
else:
l = 0
r = len(b)
while l < r - 1:
m = l + r >> 1
if b[m] < c:
l = m
else:
r = m
if b[l] < c:
l += 1
b[l] = c
return len(b)
n = int(input())
a = list(map(int, input().split()))
print(lis(a)) | FUNC_DEF ASSIGN VAR LIST FOR VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Iahub recently has learned Bubble Sort, an algorithm that is used to sort a permutation with n elements a1, a2, ..., an in ascending order. He is bored of this so simple algorithm, so he invents his own graph. The graph (let's call it G) initially has n vertices and 0 edges. During Bubble Sort execution, edges appear as described in the following algorithm (pseudocode).
procedure bubbleSortGraph()
build a graph G with n vertices and 0 edges
repeat
swapped = false
for i = 1 to n - 1 inclusive do:
if a[i] > a[i + 1] then
add an undirected edge in G between a[i] and a[i + 1]
swap( a[i], a[i + 1] )
swapped = true
end if
end for
until not swapped
/* repeat the algorithm as long as swapped value is true. */
end procedure
For a graph, an independent set is a set of vertices in a graph, no two of which are adjacent (so there are no edges between vertices of an independent set). A maximum independent set is an independent set which has maximum cardinality. Given the permutation, find the size of the maximum independent set of graph G, if we use such permutation as the premutation a in procedure bubbleSortGraph.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 105). The next line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n).
Output
Output a single integer — the answer to the problem.
Examples
Input
3
3 1 2
Output
2
Note
Consider the first example. Bubble sort swaps elements 3 and 1. We add edge (1, 3). Permutation is now [1, 3, 2]. Then bubble sort swaps elements 3 and 2. We add edge (2, 3). Permutation is now sorted. We have a graph with 3 vertices and 2 edges (1, 3) and (2, 3). Its maximal independent set is [1, 2]. | import sys
sys.setrecursionlimit(1000000)
def solve():
(n,) = rv()
(a,) = rl(1)
mem = [10000000] * (n + 1)
mem[0] = a[0]
for i in range(1, n):
left, right = 0, n - 1
while left < right:
mid = (left + right) // 2
if a[i] < mem[mid]:
right = mid
else:
left = mid + 1
mem[left] = a[i]
res = 0
for i in range(1, n):
if mem[i] != 10000000:
res = i
print(res + 1)
def rv():
return map(int, input().split())
def rl(n):
return [list(map(int, input().split())) for _ in range(n)]
if sys.hexversion == 50594544:
sys.stdin = open("test.txt")
solve() | IMPORT EXPR FUNC_CALL VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
Note
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.
<image> After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | n = int(input())
h = list(map(int, input().split()))
dp = []
dp.append(1)
for i in range(1, n - 1, 1):
dp.append(min(dp[i - 1] + 1, h[i]))
dp.append(1)
for i in range(n - 2, 0, -1):
dp[i] = min(dp[i], dp[i + 1] + 1)
mx = -1
for i in range(n):
mx = max(mx, dp[i])
print(mx) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
Note
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.
<image> After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | __author__ = "JohnHook"
n = list(map(int, input().split()))[0] + 2
a = [0] + list(map(int, input().split())) + [0]
d = [0]
for i in range(1, n):
if a[i] >= d[i - 1] + 1:
d.append(d[i - 1] + 1)
continue
d.append(a[i])
d.reverse()
a.reverse()
for i in range(1, n):
if a[i] >= d[i - 1] + 1:
d[i] = min(d[i], d[i - 1] + 1)
continue
d[i] = a[i]
print(max(d)) | ASSIGN VAR STRING ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR LIST NUMBER ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
Note
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.
<image> After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | n = int(input())
h = [0] + list(map(int, input().split()))
res = [0] * (n + 1)
Min = 0
for i in range(1, n + 1):
Min = min(Min, h[i] - i)
res[i] = i + Min
Min = n + 1
for i in range(n, 0, -1):
Min = min(Min, h[i] + i)
res[i] = min(res[i], Min - i)
ans = max(res)
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
Note
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.
<image> After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | def main():
n = int(input())
a = [int(i) for i in input().split()]
dpl = [1] * n
dpr = [1] * n
for i in range(1, n):
dpl[i] = min(dpl[i - 1] + 1, a[i])
for i in range(n - 2, -1, -1):
dpr[i] = min(dpr[i + 1] + 1, a[i])
ans = 0
for i in range(n):
ans = max(ans, min(dpl[i], dpr[i]))
print(ans)
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
Note
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.
<image> After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | n = int(input())
h = list(map(int, input().split()))
min_h = 1
for i in range(n):
if h[i] > min_h:
h[i] = min_h
else:
min_h = h[i]
min_h += 1
min_h = 1
for i in range(n - 1, -1, -1):
if h[i] > min_h:
h[i] = min_h
else:
min_h = h[i]
min_h += 1
print(max(h)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
Note
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.
<image> After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | n = int(input())
x = list(map(int, input().split()))
pre, suf, ans = [0] * n, [0] * n, 0
for i in range(n):
if i == 0:
pre[i] = 1
else:
pre[i] = min(pre[i - 1] + 1, x[i])
for i in range(n - 1, -1, -1):
if i == n - 1:
suf[i] = 1
else:
suf[i] = min(suf[i + 1] + 1, x[i])
ans = max(ans, min(pre[i], suf[i]))
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR BIN_OP LIST NUMBER VAR BIN_OP LIST NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
Note
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.
<image> After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | x = int(input())
mas = list(map(int, input().split(" ")))
mas2 = [0] * x
mas2[0] = 1
for i in range(1, x):
mas2[i] = min(mas[i], mas2[i - 1] + 1)
mas2[-1] = 1
for i in range(2, x + 1):
mas2[-i] = min(mas[-i], mas2[-i + 1] + 1, mas2[-i])
print(max(mas2)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of hi identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input
The first line contains single integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers h1, h2, ..., hn (1 ≤ hi ≤ 109) — sizes of towers.
Output
Print the number of operations needed to destroy all towers.
Examples
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
Note
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color.
<image> After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | n = int(input())
a = [int(x) for x in input().split()]
res = [0] * n
for i in range(n):
res[i] = [0, 0, 0]
if n < 3:
print(1)
else:
res[0][0] = 1
res[-1][0] = 1
for i in range(1, n - 1):
res[i][0] = min(a[i - 1] + 1, a[i + 1] + 1, a[i])
cur_min = 0
for i in range(0, n):
res[i][1] = cur_min + 1
cur_min = min(res[i][0], res[i][1])
cur_min = 0
for i in range(n - 1, -1, -1):
res[i][2] = cur_min + 1
cur_min = min(res[i][0], res[i][2])
tres = min(res[0])
for k in res:
tres = max(tres, min(k))
print(tres) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR LIST NUMBER NUMBER NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER NUMBER NUMBER ASSIGN VAR NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | def main():
n = int(input())
hts = list(map(int, input().split()))
dpleft = [1]
for i in range(1, n):
dpleft.append(min(hts[i], dpleft[-1] + 1))
hts.reverse()
dpright = [1]
for i in range(1, n):
dpright.append(min(hts[i], dpright[-1] + 1))
dpright.reverse()
ans = 0
for i in range(n):
ans = max(ans, min(dpleft[i], dpright[i]))
print(ans)
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | import sys
try:
while True:
n = int(input())
height = list(map(int, input().split(" ")))
L = [(0) for i in range(100001)]
R = [(0) for i in range(100001)]
for i in range(n):
L[i + 1] = min(L[i] + 1, height[i])
for i in range(n - 1, -1, -1):
R[i] = min(R[i + 1] + 1, height[i])
ans = 0
for i in range(1, n + 1):
ans = max(ans, min(R[i - 1], L[i]))
print(ans)
except EOFError:
pass | IMPORT WHILE NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | input()
def f(l):
c = [0] * len(l)
c[0] = c[len(c) - 1] = 9999999
for item in range(1, len(l) - 1):
c[item] = min(item, l[item], c[item - 1] + 1)
for item in reversed(range(1, len(l) - 1)):
c[item] = min(c[item], len(l) - 2 + 1 - item, c[item + 1] + 1)
return max(c[1:-1])
print(f([0] + list(map(int, input().split())) + [0])) | EXPR FUNC_CALL VAR FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER RETURN FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR LIST NUMBER |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | n = int(input())
a = [0] + [int(i) for i in input().split()] + [0]
l, r = [0] * (n + 2), [0] * (n + 2)
for i in range(1, n + 1):
l[i] = min(a[i], l[i - 1] + 1)
for i in range(n, 0, -1):
r[i] = min(a[i], r[i + 1] + 1)
ans = 0
for i in range(1, n + 1):
ans = max(ans, min(l[i], r[i]))
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP LIST NUMBER FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR LIST NUMBER ASSIGN VAR VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER BIN_OP LIST NUMBER BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | n = int(input())
a = [int(x) for x in input().split()]
a[0] = 1
a[-1] = 1
for i in range(1, n):
a[i] = min(a[i], a[i - 1] + 1)
a[-i] = min(a[-i], a[-(i - 1)] + 1)
print(max(a)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | from sys import stdin, stdout
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int, stdin.readline().split()))
PI = float("inf")
for _ in range(1):
n = nmbr()
a = lst()
ans = 0
pre = [0] * n
suf = [0] * n
pre[0] = suf[n - 1] = 1
for i in range(1, n):
pre[i] = min(1 + pre[i - 1], a[i])
for i in range(n - 2, -1, -1):
suf[i] = min(1 + suf[i + 1], a[i])
for pp, ss in zip(pre, suf):
ans = max(ans, min(pp, ss))
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP VAR NUMBER VAR VAR FOR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | input()
l, r = [0], [0]
for a, b in (lambda a: list(zip(a, reversed(a))))(
[0] + list(map(int, input().split())) + [0]
):
l.append(min(a, l[-1] + 1))
r.append(min(b, r[-1] + 1))
print(max(list(map(min, list(zip(l[1:], reversed(r[1:]))))))) | EXPR FUNC_CALL VAR ASSIGN VAR VAR LIST NUMBER LIST NUMBER FOR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR LIST NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | n = int(input())
ar = list(map(int, input().split()))
r = []
l = []
day = 0
for i in ar:
day += 1
day = min(i, day)
l.append(day)
day = 0
for i in reversed(ar):
day += 1
day = min(i, day)
r.append(day)
ans = 0
x = 0
kk = n - 1
while x < n:
ans = max(ans, min(r[kk], l[x]))
x += 1
kk -= 1
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | x = int(input())
y = list(map(int, input().split(" ")))
y[0] = 1
y[x - 1] = 1
z = y[:]
for i in range(1, x):
z[i] = min(z[i], z[i - 1] + 1)
w = y[:]
for i in range(x - 2, -1, -1):
w[i] = min(w[i], w[i + 1] + 1)
ans = 0
for i in range(x):
ans = max(ans, min(z[i], w[i]))
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | def powmod(x, p, m):
if p <= 0:
return 1
if p <= 1:
return x % m
return powmod(x * x % m, p // 2, m) * (x % m) ** (p % 2) % m
n = int(input())
h = [0] + [int(x) for x in input().split()] + [0]
for i in range(1, len(h)):
h[i] = min(h[i], h[i - 1] + 1)
h[-i - 1] = min(h[-i - 1], h[-i] + 1)
ans = max(h)
print(ans) | FUNC_DEF IF VAR NUMBER RETURN NUMBER IF VAR NUMBER RETURN BIN_OP VAR VAR RETURN BIN_OP BIN_OP FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP LIST NUMBER FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | R = lambda: map(int, input().split())
n = int(input())
arr = [-1] + list(R()) + [-1]
dpl, dpr = arr[:], arr[:]
for i in range(1, n + 1):
dpl[i] = max(0, min(dpl[i - 1] + 1, arr[i] - 1))
for i in range(n, 0, -1):
dpr[i] = max(0, min(dpr[i + 1] + 1, arr[i] - 1))
print(1 + max(min(l, r) for l, r in zip(dpl, dpr))) | ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP LIST NUMBER FUNC_CALL VAR FUNC_CALL VAR LIST NUMBER ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR |
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built n towers in a row. The i-th tower is made of h_{i} identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
-----Input-----
The first line contains single integer n (1 ≤ n ≤ 10^5).
The second line contains n space-separated integers h_1, h_2, ..., h_{n} (1 ≤ h_{i} ≤ 10^9) — sizes of towers.
-----Output-----
Print the number of operations needed to destroy all towers.
-----Examples-----
Input
6
2 1 4 6 2 2
Output
3
Input
7
3 3 3 1 3 3 3
Output
2
-----Note-----
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. [Image] After first operation there are four blocks left and only one remains after second operation. This last block is destroyed in third operation. | def main():
input()
hh = list(map(int, input().split()))
for f in (True, False):
m = 1
for i, h in enumerate(hh):
if h > m:
hh[i] = m
else:
m = h
m += 1
if f:
hh.reverse()
print(max(hh))
main() | FUNC_DEF EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FOR VAR NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | def bfs(arr, i, d, memo):
if i in memo:
return memo[i]
res = 1
for j in range(max(i - 1, 0), max(i - d - 1, -1), -1):
if arr[j] >= arr[i]:
break
res = max(res, bfs(arr, j, d, memo) + 1)
for j in range(i + 1, min(i + d + 1, len(arr))):
if arr[j] >= arr[i]:
break
res = max(res, bfs(arr, j, d, memo) + 1)
memo[i] = res
return res
class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
memo = dict()
return max(bfs(arr, i, d, memo) for i in range(len(arr))) | FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER NUMBER IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
memo = {}
def solve(i):
if i in memo:
return memo[i]
ret = 1
j = i - 1
while j >= max(0, i - d) and arr[j] < arr[i]:
ret = max(ret, 1 + solve(j))
j -= 1
j = i + 1
while j <= min(len(arr) - 1, i + d) and arr[j] < arr[i]:
ret = max(ret, 1 + solve(j))
j += 1
memo[i] = ret
return ret
for i in range(len(arr)):
solve(i)
return max(memo.values()) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR DICT FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, nums: List[int], d: int) -> int:
nums.append(math.inf)
dp = [1] * len(nums)
stack = []
for i, cur in enumerate(nums):
while stack and nums[stack[-1]] < cur:
poped_same_val = [stack.pop()]
while stack and nums[stack[-1]] == nums[poped_same_val[0]]:
poped_same_val.append(stack.pop())
for poped in poped_same_val:
if i - poped <= d:
dp[i] = max(dp[i], dp[poped] + 1)
if stack and poped - stack[-1] <= d:
left_first_big_of_poped = stack[-1]
dp[left_first_big_of_poped] = max(
dp[left_first_big_of_poped], dp[poped] + 1
)
stack.append(i)
return max(dp[:-1]) | CLASS_DEF FUNC_DEF VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR NUMBER VAR ASSIGN VAR LIST FUNC_CALL VAR WHILE VAR VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR NUMBER VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
book = [1] * len(arr)
index = collections.defaultdict(list)
for i in range(len(arr)):
index[arr[i]].append(i)
val = list(index.keys())
val.sort(reverse=True)
for v in val:
for i in index[v]:
height = arr[i]
for j in range(i + 1, i + d + 1):
if j == len(arr):
break
if arr[j] > height:
book[i] = max(book[i], book[j] + 1)
height = arr[j]
height = arr[i]
for j in range(i - 1, i - d - 1, -1):
if j == -1:
break
if arr[j] > height:
book[i] = max(book[i], book[j] + 1)
height = arr[j]
return max(book) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER FOR VAR VAR FOR VAR VAR VAR ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def rec(self, v):
if self.memo[v] != -1:
return self.memo[v]
res = 1
for nv in self.G[v]:
res = max(res, 1 + self.rec(nv))
self.memo[v] = res
return res
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
self.G = [[] for _ in range(n)]
for i in range(n):
for j in range(i - 1, i - d - 1, -1):
if j < 0 or arr[j] >= arr[i]:
break
self.G[i].append(j)
for j in range(i + 1, i + d + 1):
if j >= n or arr[j] >= arr[i]:
break
self.G[i].append(j)
self.memo = [-1] * n
ans = 0
for i in range(n):
ans = max(ans, self.rec(i))
return ans | CLASS_DEF FUNC_DEF IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR RETURN VAR FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
def helper(arr, index, d, memo):
if arr[index] < 0 or not 0 <= index < len(arr):
return 0
if index in memo:
return memo[index]
left_visited = 0
right_visited = 0
arr[index] = -arr[index]
for i in range(index + 1, index + d + 1):
if i >= len(arr):
break
if abs(arr[i]) >= abs(arr[index]):
break
if arr[i] < 0:
continue
left_visited = max(left_visited, helper(arr, i, d, memo))
for j in range(index - 1, index - d - 1, -1):
if j < 0:
break
if abs(arr[j]) >= abs(arr[index]):
break
if arr[j] < 0:
continue
right_visited = max(right_visited, helper(arr, j, d, memo))
arr[index] = -arr[index]
memo[index] = 1 + max(left_visited, right_visited)
return memo[index]
max_visited = 1
p = False
memo = {}
for index in range(len(arr)):
visited = helper(arr, index, d, memo)
max_visited = max(visited, max_visited)
return max_visited | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR VAR RETURN NUMBER IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
i2h = {}
for i, h in enumerate(arr):
i2h[i] = h, i
dp = [1] * n
for h, idx in sorted(i2h.values(), key=lambda hi: hi[0]):
j = idx + 1
while j <= min(n - 1, idx + d) and arr[idx] > arr[j]:
dp[idx] = max(dp[idx], dp[j] + 1)
j += 1
j = idx - 1
while j >= max(0, idx - d) and arr[idx] > arr[j]:
dp[idx] = max(dp[idx], dp[j] + 1)
j -= 1
return max(dp)
def maxJumps__(self, arr: List[int], d: int) -> int:
n = len(arr)
memo = [0] * n
def dfs(idx):
if memo[idx]:
return memo[idx]
jump = 1
j = idx + 1
while j <= min(n - 1, idx + d) and arr[idx] > arr[j]:
jump = max(jump, dfs(j) + 1)
j += 1
j = idx - 1
while j >= max(0, idx - d) and arr[idx] > arr[j]:
jump = max(jump, dfs(j) + 1)
j -= 1
memo[idx] = jump
return jump
max_val = 0
for i in range(n):
max_val = max(max_val, dfs(i))
return max_val
def maxJumps_(self, arr: List[int], d: int) -> int:
def backtracking(idx):
nonlocal memo, max_val
if idx in memo:
return memo[idx]
lstack = []
rstack = []
lbarrier = False
rbarrier = False
for i in range(1, d + 1):
if not lbarrier and idx - i >= 0 and arr[idx - i] < arr[idx]:
lstack.append(idx - i)
else:
lbarrier = True
if not rbarrier and idx + i < len(arr) and arr[idx + i] < arr[idx]:
rstack.append(idx + i)
else:
rbarrier = True
ljump = 1
while lstack:
lidx = lstack.pop()
ljump = max(ljump, 1 + backtracking(lidx))
rjump = 1
while rstack:
ridx = rstack.pop()
rjump = max(rjump, 1 + backtracking(ridx))
jump = max(ljump, rjump)
memo[idx] = jump
max_val = max(max_val, jump)
return jump
memo = {}
max_val = 0
for i in range(len(arr)):
backtracking(i)
return max_val | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR DICT FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER RETURN FUNC_CALL VAR VAR VAR FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
@lru_cache(None)
def helper(idx):
ans = 1
ub, lb = 1, 1
while ub <= d or lb <= d:
if idx + ub < len(arr) and arr[idx + ub] < arr[idx]:
ans = max(ans, 1 + helper(idx + ub))
ub += 1
else:
ub = sys.maxsize
if idx - lb >= 0 and arr[idx - lb] < arr[idx]:
ans = max(ans, 1 + helper(idx - lb))
lb += 1
else:
lb = sys.maxsize
return ans
return max([helper(idx) for idx in range(len(arr))]) | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER WHILE VAR VAR VAR VAR IF BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR RETURN VAR FUNC_CALL VAR NONE RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, A: List[int], d: int) -> int:
n = len(A)
@lru_cache(None)
def helper(i):
ans = 1
l = max(0, i - d)
r = min(n - 1, i + d)
for j in reversed(range(l, i)):
if A[j] < A[i]:
ans = max(ans, 1 + helper(j))
else:
break
for j in range(i + 1, r + 1):
if A[j] < A[i]:
ans = max(ans, 1 + helper(j))
else:
break
return ans
ans = max([helper(i) for i in range(n)])
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR RETURN VAR FUNC_CALL VAR NONE ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
maxPath = 0
memory = {i: (0) for i in range(len(arr))}
for i in range(len(arr)):
maxPath = max(maxPath, self.dfs(i, d, arr, len(arr), memory))
return maxPath
def dfs(self, index, d, arr, arrLen, memory):
if memory[index] != 0:
return memory[index]
currentMax = 0
for i in range(index + 1, index + d + 1):
if i >= arrLen or arr[index] <= arr[i]:
break
currentMax = max(currentMax, self.dfs(i, d, arr, arrLen, memory))
for i in range(index - 1, index - d - 1, -1):
if i < 0 or arr[index] <= arr[i]:
break
currentMax = max(currentMax, self.dfs(i, d, arr, arrLen, memory))
memory[index] = currentMax + 1
return memory[index] | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR FUNC_DEF IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
N = len(arr)
dp = [None] * N
def cdp(i):
if dp[i] != None:
return dp[i]
dp[i] = 1
for j in range(i - 1, max(0, i - d) - 1, -1):
if arr[j] >= arr[i]:
break
dp[i] = max(dp[i], 1 + cdp(j))
for j in range(i + 1, min(N - 1, i + d) + 1):
if arr[j] >= arr[i]:
break
dp[i] = max(dp[i], 1 + cdp(j))
return dp[i]
for i in range(N):
cdp(i)
return max(dp) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NONE VAR FUNC_DEF IF VAR VAR NONE RETURN VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR NUMBER BIN_OP VAR VAR NUMBER NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR RETURN VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
mem = [0] * len(arr)
def dfs(i):
if mem[i]:
return mem[i]
num = 1
j = i + 1
while j in range(i + 1, min(len(arr), i + d + 1)) and arr[j] < arr[i]:
num = max(num, dfs(j) + 1)
j += 1
j = i - 1
while j in range(max(0, i - d), i) and arr[j] < arr[i]:
num = max(num, dfs(j) + 1)
j -= 1
mem[i] = num
return mem[i]
res = 0
for i in range(len(arr)):
dfs(i)
res = max(res, dfs(i) + 1)
return max(mem) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, A: List[int], d: int) -> int:
dp = [1] * len(A)
steps = [(v, i) for i, v in enumerate(A)]
steps.sort()
for val, i in steps:
maxVal = val
for j in range(i + 1, min(i + d + 1, len(steps))):
if A[j] > maxVal:
maxVal = max(maxVal, A[j])
dp[j] = max(dp[j], dp[i] + 1)
maxVal = val
for j in reversed(range(max(i - d, 0), i)):
if A[j] > maxVal:
maxVal = max(maxVal, A[j])
dp[j] = max(dp[j], dp[i] + 1)
return max(dp) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FOR VAR VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR IF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
visited = {}
arrLen = len(arr)
def check(i0):
if i0 in visited:
return visited[i0]
di = 1
preV = 0
subAns = 1
while di <= d and i0 + di < arrLen and arr[i0 + di] < arr[i0]:
if arr[i0 + di] >= preV:
subAns = max(subAns, 1 + check(i0 + di))
preV = max(preV, arr[i0 + di])
di += 1
di = 1
preV = 0
while di <= d and i0 - di >= 0 and arr[i0 - di] < arr[i0]:
if arr[i0 - di] >= preV:
subAns = max(subAns, 1 + check(i0 - di))
preV = max(preV, arr[i0 - di])
di += 1
visited[i0] = subAns
return subAns
ans = 1
for i in range(arrLen):
ans = max(ans, check(i))
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR DICT ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR IF VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR IF VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], dis: int) -> int:
res = 1
d = dict()
visit = set()
if len(arr) == 1:
return 1
for i in range(len(arr)):
if i == 0:
if arr[i] <= arr[i + 1]:
d[i] = 1
visit.add(i)
elif i == len(arr) - 1:
if arr[i] <= arr[i - 1]:
d[i] = 1
visit.add(i)
elif arr[i] <= arr[i - 1] and arr[i] <= arr[i + 1]:
d[i] = 1
visit.add(i)
def dfs(index):
if index not in visit:
visit.add(index)
cur = 1
for i in range(1, dis + 1):
if index - i >= 0 and arr[index - i] < arr[index]:
temp = dfs(index - i)
d[index - i] = temp
cur = max(cur, 1 + temp)
else:
break
for i in range(1, dis + 1):
if index + i < len(arr) and arr[index + i] < arr[index]:
temp = dfs(index + i)
d[index + i] = temp
cur = max(cur, 1 + temp)
else:
break
return cur
else:
return d[index]
for x in range(len(arr)):
if x not in visit:
cur = dfs(x)
d[x] = cur
res = max(res, cur)
else:
res = max(res, d[x])
return res | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER RETURN NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR RETURN VAR RETURN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, A: List[int], d: int) -> int:
n = len(A)
dp = [1] * n
for a, i in sorted([a, i] for i, a in enumerate(A)):
j = i - 1
while j >= 0 and A[j] < A[i] and i - j <= d:
dp[i] = max(dp[i], dp[j] + 1)
j -= 1
j = i + 1
while j < n and A[j] < A[i] and j - i <= d:
dp[i] = max(dp[i], dp[j] + 1)
j += 1
return max(dp) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
dp = [(0) for i in range(len(arr))]
def findMaxJumps(i):
if dp[i] != 0:
return dp[i]
else:
ans = 1
for j in range(i + 1, min(i + d + 1, len(arr))):
if arr[j] >= arr[i]:
break
ans = max(ans, 1 + findMaxJumps(j))
for j in range(i - 1, max(i - d, 0) - 1, -1):
if arr[j] >= arr[i]:
break
ans = max(ans, 1 + findMaxJumps(j))
dp[i] = ans
return ans
ans = 1
for i in range(len(arr)):
ans = max(ans, findMaxJumps(i))
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER NUMBER NUMBER IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
ans = [1] * len(arr)
def dp(i):
nonlocal ans, arr, d
if ans[i] != 1:
return ans[i]
for dr in [-1, 1]:
for k in range(i, i + d * dr + dr, dr):
if k < 0 or k == i:
continue
if k > len(arr) - 1 or arr[k] >= arr[i]:
break
ans[i] = max(ans[i], 1 + dp(k))
return ans[i]
for i in range(len(arr)):
dp(i)
return max(ans) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR NUMBER RETURN VAR VAR FOR VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR RETURN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution2:
def maxJumps(self, arr, d):
n = len(arr)
dp = [1] * n
for a, i in sorted([a, i] for i, a in enumerate(arr)):
for di in [-1, 1]:
for j in range(i + di, i + di + d * di, di):
if not (0 <= j < n and arr[j] < arr[i]):
break
dp[i] = max(dp[i], dp[j] + 1)
return max(dp)
class Solution:
def maxJumps(self, arr, d):
n = len(arr)
dp = [1] * (n + 1)
stack = []
for i, a in enumerate(arr + [float("inf")]):
while stack and arr[stack[-1]] < a:
L = [stack.pop()]
while stack and arr[stack[-1]] == arr[L[0]]:
L.append(stack.pop())
for j in L:
if i - j <= d:
dp[i] = max(dp[i], dp[j] + 1)
if stack and j - stack[-1] <= d:
dp[stack[-1]] = max(dp[stack[-1]], dp[j] + 1)
stack.append(i)
return max(dp[:-1])
class Solution1:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
res = [0] * n
def dp(i):
if res[i]:
return res[i]
res[i] = 1
for di in [-1, 1]:
for j in range(i + di, i + di + d * di, di):
if not (0 <= j < n and arr[j] < arr[i]):
break
res[i] = max(res[i], dp(j) + 1)
return res[i]
return max(list(map(dp, list(range(n))))) | CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR FOR VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR VAR IF NUMBER VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER RETURN FUNC_CALL VAR VAR CLASS_DEF FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER BIN_OP VAR NUMBER ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR BIN_OP VAR LIST FUNC_CALL VAR STRING WHILE VAR VAR VAR NUMBER VAR ASSIGN VAR LIST FUNC_CALL VAR WHILE VAR VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FOR VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER FUNC_CALL VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR NUMBER CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR VAR NUMBER FOR VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR VAR IF NUMBER VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
max_less_than = []
def jump(iter):
res = [None for i in range(len(arr))]
stack = deque([])
for i in iter:
while stack and abs(i - stack[0]) > d:
stack.popleft()
while stack and arr[i] > arr[stack[-1]]:
latest_poped = stack.pop()
res[latest_poped] = i
stack.append(i)
return res
max_less_than.append(jump(range(len(arr))))
max_less_than.append(jump(reversed(range(len(arr)))))
@lru_cache(None)
def dp(i):
if i is None:
return 0
return (
max(map(dp, [max_less_than[0][i], max_less_than[1][i]]), default=0) + 1
)
return max(map(dp, range(len(arr)))) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR LIST FUNC_DEF ASSIGN VAR NONE VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR LIST FOR VAR VAR WHILE VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR WHILE VAR VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR NONE RETURN NUMBER RETURN BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR LIST VAR NUMBER VAR VAR NUMBER VAR NUMBER NUMBER FUNC_CALL VAR NONE RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
dp = [(1) for i in range(n)]
p = []
for idx, x in enumerate(arr):
p.append((x, idx))
p.sort(reverse=True)
for i in range(n):
idx = p[i][1]
for j in range(idx + 1, min(n, d + idx + 1)):
if arr[j] >= arr[idx]:
break
dp[j] = max(dp[j], dp[idx] + 1)
for j in range(idx - 1, max(-1, idx - d - 1), -1):
if arr[j] >= arr[idx]:
break
dp[j] = max(dp[j], dp[idx] + 1)
return max([dp[i] for i in range(n)]) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
def find_longest_l(pos, d):
if pos - 1 < 0 or arr[pos - 1] > arr[pos]:
return pos
i = pos
while i > 0 and abs(pos - i) < d and arr[i - 1] < arr[pos]:
i -= 1
return i
def find_longest_r(pos, d):
if pos + 1 >= len(arr) or arr[pos + 1] > arr[pos]:
return pos
i = pos
while i < len(arr) - 1 and abs(pos - i) < d and arr[i + 1] < arr[pos]:
i += 1
return i
def jump(pos, number_of_locations, seen):
maximum_locations = number_of_locations
L = find_longest_l(pos, d)
R = find_longest_r(pos, d)
for i in range(L, R + 1):
if i == pos:
continue
if i not in seen:
jumps_to_i = jump(i, number_of_locations, seen)
maximum_locations = max(maximum_locations, jumps_to_i + 1)
else:
jumps_to_i = seen[i]
maximum_locations = max(maximum_locations, jumps_to_i + 1)
seen[pos] = maximum_locations
return maximum_locations
indices = [(i, arr[i]) for i in range(len(arr))]
indices.sort(key=lambda x: x[1], reverse=True)
seen = {}
return max(jump(x[0], 1, seen) for x in indices) | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR VAR RETURN VAR ASSIGN VAR VAR WHILE VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER RETURN VAR FUNC_DEF IF BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER VAR VAR RETURN VAR ASSIGN VAR VAR WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR DICT RETURN FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER NUMBER VAR VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
def backtracking(idx):
nonlocal memo, max_val
if idx in memo:
return memo[idx]
lstack = []
rstack = []
lbarrier = False
rbarrier = False
for i in range(1, d + 1):
if not lbarrier and idx - i >= 0 and arr[idx - i] < arr[idx]:
lstack.append(idx - i)
else:
lbarrier = True
if not rbarrier and idx + i < len(arr) and arr[idx + i] < arr[idx]:
rstack.append(idx + i)
else:
rbarrier = True
ljump = 1
while lstack:
lidx = lstack.pop()
ljump = max(ljump, 1 + backtracking(lidx))
rjump = 1
while rstack:
ridx = rstack.pop()
rjump = max(rjump, 1 + backtracking(ridx))
jump = max(ljump, rjump)
memo[idx] = jump
max_val = max(max_val, jump)
return jump
memo = {}
max_val = 0
for i in range(len(arr)):
backtracking(i)
return max_val | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR RETURN VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
N = len(arr)
@lru_cache(None)
def rec(idx):
if not 0 <= idx < N:
return -inf
ans = 1
for i in range(idx - 1, idx - d - 1, -1):
if i < 0 or arr[i] >= arr[idx]:
break
ans = max(ans, rec(i) + 1)
for i in range(idx + 1, idx + d + 1):
if i >= N or arr[i] >= arr[idx]:
break
ans = max(ans, rec(i) + 1)
return ans
ans = 0
for i in range(N):
ans = max(ans, rec(i))
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF IF NUMBER VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR FUNC_CALL VAR NONE ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
if n <= 1:
return n
visited = [(False) for i in range(n)]
L_neighbor = {}
R_neighbor = {}
for i in range(n):
L_neighbor[i] = i
R_neighbor[i] = i
dec = deque()
dec.append(0)
for i in range(1, n):
while len(dec) > 0:
if dec[0] < i - d:
dec.popleft()
else:
break
while len(dec) > 0:
if arr[dec[len(dec) - 1]] < arr[i]:
dec.pop()
else:
break
if len(dec) == 0:
L_neighbor[i] = max(i - d, 0)
dec.append(i)
else:
L_neighbor[i] = max(dec[len(dec) - 1] + 1, i - d, 0)
dec.append(i)
dec = deque()
dec.append(n - 1)
for i in range(n - 2, -1, -1):
while len(dec) > 0:
if dec[0] > i + d:
dec.popleft()
else:
break
while len(dec) > 0:
if arr[dec[len(dec) - 1]] < arr[i]:
dec.pop()
else:
break
if len(dec) == 0:
R_neighbor[i] = min(i + d, n - 1)
dec.append(i)
else:
R_neighbor[i] = min(dec[len(dec) - 1] - 1, i + d, n - 1)
dec.append(i)
res = 0
limit = [(0) for i in range(n)]
for i in range(n):
self.maxJumpDFS(i, arr, L_neighbor, R_neighbor, visited, limit)
for i in range(n):
if res < limit[i]:
res = limit[i]
return res
def maxJumpDFS(self, start, arr, L_neighbor, R_neighbor, visited, limit):
if visited[start]:
return limit[start]
else:
currmax = 0
for i in range(L_neighbor[start], R_neighbor[start] + 1):
if i == start:
continue
curr = self.maxJumpDFS(i, arr, L_neighbor, R_neighbor, visited, limit)
if currmax < curr:
currmax = curr
visited[start] = True
limit[start] = currmax + 1
return limit[start] | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER RETURN VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR WHILE FUNC_CALL VAR VAR NUMBER IF VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR NUMBER IF VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER WHILE FUNC_CALL VAR VAR NUMBER IF VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR WHILE FUNC_CALL VAR VAR NUMBER IF VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR RETURN VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
dp = [(-1) for _ in range(len(arr))]
def dp_memo(x):
if dp[x] != -1:
return dp[x]
dp[x] = 1
for y in range(x - 1, x - d - 1, -1):
if y < 0 or arr[y] >= arr[x]:
break
dp[x] = max(dp[x], 1 + dp_memo(y))
for y in range(x + 1, x + d + 1, +1):
if y >= len(arr) or arr[y] >= arr[x]:
break
dp[x] = max(dp[x], 1 + dp_memo(y))
return dp[x]
for i in range(len(arr)):
dp_memo(i)
print(dp)
return max(dp) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR RETURN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
dp = [(1) for i in range(n)]
@lru_cache(None)
def rec(i):
j = 1
while j <= d and i - j >= 0 and arr[i - j] < arr[i]:
dp[i] = max(dp[i], 1 + rec(i - j))
j += 1
j = 1
while j <= d and i + j < n and arr[i + j] < arr[i]:
dp[i] = max(dp[i], 1 + rec(i + j))
j += 1
return dp[i]
for k in range(n):
rec(k)
return max(dp) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR NUMBER WHILE VAR VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER RETURN VAR VAR FUNC_CALL VAR NONE FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
dp = [1] * n
for a, i in sorted([a, i] for i, a in enumerate(arr)):
for di in [-1, 1]:
for j in range(i + di, i + d * di + di, di):
if not (0 <= j < n and arr[j] < arr[i]):
break
dp[i] = max(dp[i], dp[j] + 1)
return max(dp) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR FOR VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR VAR IF NUMBER VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
def helper(i):
if cache[i]:
return cache[i]
numberOfJump = 0
for j in range(i + 1, i + d + 1):
if j >= len(arr) or arr[j] >= arr[i]:
break
numberOfJump = max(helper(j), numberOfJump)
for j in range(i - 1, i - d - 1, -1):
if j < 0 or arr[j] >= arr[i]:
break
numberOfJump = max(helper(j), numberOfJump)
cache[i] = 1 + numberOfJump
return cache[i]
cache = [0] * len(arr)
ans = 0
for i in range(len(arr)):
ans = max(helper(i), ans)
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR BIN_OP NUMBER VAR RETURN VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
graph = collections.defaultdict(list)
st = []
for i in range(n):
while st and arr[st[-1]] < arr[i]:
j = st.pop()
if i - j <= d:
graph[j].append(i)
st.append(i)
st = []
for i in range(n - 1, -1, -1):
while st and arr[st[-1]] < arr[i]:
j = st.pop()
if j - i <= d:
graph[j].append(i)
st.append(i)
visited = {}
def dfs(i):
if i in visited:
return visited[i]
step = 1
for j in graph[i]:
step = max(step, 1 + dfs(j))
visited[i] = step
return step
res = 0
for i in range(n):
res = max(res, dfs(i))
return res | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER WHILE VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR DICT FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
dp = {}
def jump(i):
if i in dp:
return dp[i]
m = 1
step = 1
while i + step < len(arr) and step <= d:
if arr[i] <= arr[i + step]:
break
m = max(m, jump(i + step) + 1)
step += 1
step = 1
while i - step >= 0 and step <= d:
if arr[i] <= arr[i - step]:
break
m = max(m, jump(i - step) + 1)
step += 1
dp[i] = m
return m
ans = 0
for i in range(len(arr)):
if i not in dp:
jump(i)
ans = max(ans, dp[i])
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR DICT FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR NUMBER VAR VAR IF VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
@lru_cache(None)
def jump(mid):
left = max(mid - d, 0)
right = min(mid + d, len(arr) - 1)
res = 1
for i in range(mid - 1, left - 1, -1):
if arr[i] >= arr[mid]:
break
res = max(res, jump(i) + 1)
for i in range(mid + 1, right + 1):
if arr[i] >= arr[mid]:
break
res = max(res, jump(i) + 1)
return res
res = [1] * len(arr)
for i in range(len(arr)):
res[i] = jump(i)
return max(res) | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR FUNC_CALL VAR NONE ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
l = len(arr)
self.check, self.seen = [[] for _ in range(l)], dict()
for i, v in enumerate(arr):
for sign in (1, -1):
for x in range(sign, (d + 1) * sign, sign):
if i + x not in list(range(l)) or v <= arr[i + x]:
break
self.check[i].append(i + x)
if not self.check[i]:
self.seen[i] = 1
return max(self.helper(i) for i in range(l))
def helper(self, p):
if p not in self.seen:
res = 0
for n in self.check[p]:
res = max(res, self.seen.get(n, self.helper(n)) + 1)
self.seen[p] = res
return self.seen[p] | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR LIST VAR FUNC_CALL VAR VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR FOR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER VAR VAR IF BIN_OP VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_DEF IF VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
@lru_cache(None)
def dfs(i):
if i == 0 and i + 1 < len(arr):
if arr[i] <= arr[i + 1]:
return 1
elif i == len(arr) - 1 and i - 1 >= 0:
if arr[i] <= arr[i - 1]:
return 1
elif (
i - 1 >= 0
and i + 1 < len(arr)
and arr[i] <= arr[i - 1]
and arr[i] <= arr[i + 1]
):
return 1
left_step = 1
for j in range(i + 1, i + d + 1):
if j >= len(arr) or arr[j] >= arr[i]:
break
left_step = max(left_step, 1 + dfs(j))
right_step = 1
for j in range(i - 1, i - d - 1, -1):
if j < 0 or arr[j] >= arr[i]:
break
right_step = max(right_step, 1 + dfs(j))
return max(left_step, right_step)
max_step = 0
for i in range(len(arr)):
max_step = max(max_step, dfs(i))
return max_step | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER RETURN NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER RETURN NUMBER IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER RETURN NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR FUNC_CALL VAR NONE ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, A: List[int], d: int) -> int:
N = len(A)
jumpable = collections.defaultdict(list)
def find_jumpable_indices(iter):
stack = []
for i in iter:
while stack and A[stack[-1]] < A[i]:
j = stack.pop()
if abs(i - j) <= d:
jumpable[i].append(j)
stack.append(i)
find_jumpable_indices(list(range(N)))
find_jumpable_indices(reversed(list(range(N))))
dp = [-1] * N
def dfs(idx):
if dp[idx] > -1:
return dp[idx]
res = 1
for j in jumpable[idx]:
res = max(res, 1 + dfs(j))
dp[idx] = res
return res
return max(list(map(dfs, list(range(N))))) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR LIST FOR VAR VAR WHILE VAR VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FUNC_DEF IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR RETURN VAR RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
N = len(arr)
umap = [list() for i in range(N)]
dp = [(-1) for i in range(N)]
def dfs(idx: int) -> int:
if idx >= N or idx < 0:
return 0
if dp[idx] != -1:
return dp[idx]
curr = 0
for u in umap[idx]:
curr = max(curr, dfs(u))
dp[idx] = curr + 1
return dp[idx]
for i in range(N):
minIdx = max(0, i - d)
maxIdx = min(N - 1, i + d)
for t in range(i - 1, minIdx - 1, -1):
if arr[i] > arr[t]:
umap[i].append(t)
else:
break
for t in range(i + 1, maxIdx + 1):
if arr[i] > arr[t]:
umap[i].append(t)
else:
break
result = 0
for i in range(N):
result = max(result, dfs(i))
return result | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FUNC_DEF VAR IF VAR VAR VAR NUMBER RETURN NUMBER IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
arr1 = [[i, v] for i, v in enumerate(arr)]
arr1.sort(key=lambda x: x[1])
dp = [0] * len(arr)
def dfs(start, arr, dp):
if dp[start] != 0:
return dp[start]
dp[start] = 1
for i in range(start - 1, max(start - d - 1, -1), -1):
if arr[start] <= arr[i]:
break
dp[start] = max(dp[start], dfs(i, arr, dp) + 1)
for i in range(start + 1, min(start + d + 1, len(arr))):
if arr[start] <= arr[i]:
break
dp[start] = max(dp[start], dfs(i, arr, dp) + 1)
return dp[start]
for i in range(len(arr1)):
dfs(arr1[i][0], arr, dp)
return max(dp) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR LIST VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER NUMBER NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER RETURN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER VAR VAR RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
ans = dict()
def dfs(idx):
if idx in ans:
return
ans[idx] = 1
i = idx - 1
while i >= 0 and idx - i <= d and arr[i] < arr[idx]:
dfs(i)
ans[idx] = max(ans[idx], ans[i] + 1)
i -= 1
i = idx + 1
while i < len(arr) and i - idx <= d and arr[i] < arr[idx]:
dfs(i)
ans[idx] = max(ans[idx], ans[i] + 1)
i += 1
for i in range(len(arr)):
dfs(i)
print(ans)
return max(ans.values()) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF VAR VAR RETURN ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER BIN_OP VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
a = []
jump = [0] * len(arr)
result = 0
for i in range(len(arr)):
a.append((i, arr[i]))
a.sort(key=lambda x: x[1])
for i in range(len(arr)):
l = a[i][0] - 1
r = a[i][0] + 1
m = 0
while l >= 0 and arr[l] < a[i][1] and a[i][0] - l <= d:
m = max(m, jump[l])
l -= 1
while r < len(arr) and arr[r] < a[i][1] and r - a[i][0] <= d:
m = max(m, jump[r])
r += 1
jump[a[i][0]] = m + 1
result = max(result, m + 1)
return result | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR VAR NUMBER BIN_OP VAR VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
memo = [-1] * len(arr)
def dfs(i):
if i < 0 or i >= len(arr):
return 0
if memo[i] > 0:
return memo[i]
memo[i] = 1
for j in reversed(list(range(max(0, i - d), i))):
if arr[j] >= arr[i]:
break
memo[i] = max(memo[i], dfs(j) + 1)
for j in range(i + 1, min(len(arr), i + d + 1)):
if arr[j] >= arr[i]:
break
memo[i] = max(memo[i], dfs(j) + 1)
return memo[i]
return max(dfs(i) for i in range(len(arr))) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FUNC_DEF IF VAR NUMBER VAR FUNC_CALL VAR VAR RETURN NUMBER IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP VAR VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
max_pos = 1
memo = {}
def helper(i):
if i in list(memo.keys()):
return memo[i]
res = 0
j = i - 1
while 0 <= j < len(arr) and i - j <= d and arr[j] < arr[i]:
res = max(res, helper(j))
j -= 1
j = i + 1
while 0 <= j < len(arr) and j - i <= d and arr[j] < arr[i]:
res = max(res, helper(j))
j += 1
res = res + 1
memo[i] = res
return res
for i in range(len(arr)):
max_pos = max(max_pos, helper(i))
return max_pos | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR DICT FUNC_DEF IF VAR FUNC_CALL VAR FUNC_CALL VAR RETURN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
def dfs(i):
if i in cache:
return cache[i]
cache[i] = 1
for neigh in g[i]:
cache[i] = max(cache[i], 1 + dfs(neigh))
return cache[i]
g = defaultdict(set)
for i in range(len(arr)):
for x in range(1, d + 1):
if i - x >= 0 and arr[i - x] < arr[i]:
g[i].add(i - x)
else:
break
for x in range(1, d + 1):
if i + x < len(arr) and arr[i + x] < arr[i]:
g[i].add(i + x)
else:
break
cache = {}
for i in range(len(arr)):
dfs(i)
return max(cache.values()) | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP NUMBER FUNC_CALL VAR VAR RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
self.memo = {}
self.result = 0
for i in range(len(arr)):
self.dfs(arr, i, d)
return self.result
def dfs(self, arr, index, d):
if index in self.memo:
return self.memo[index]
result = 0
for dir in [-1, 1]:
for i in range(1, d + 1):
j = index + i * dir
if 0 <= j < len(arr) and arr[index] > arr[j]:
result = max(result, self.dfs(arr, j, d))
else:
break
self.memo[index] = result + 1
self.result = max(self.result, result + 1)
return result + 1 | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF NUMBER VAR FUNC_CALL VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER RETURN BIN_OP VAR NUMBER |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
cache = {}
def find_max_jumps(i):
if i in cache:
return cache[i]
max_jumps = 1
for dr in [1, -1]:
for r in range(1, d + 1):
if not (0 <= i + dr * r < n and arr[i + dr * r] < arr[i]):
break
max_jumps = max(max_jumps, 1 + find_max_jumps(i + dr * r))
cache[i] = max_jumps
return max_jumps
result = 1
for i in range(n):
result = max(result, find_max_jumps(i))
return result
class Solution111:
def maxJumps(self, arr: List[int], d: int) -> int:
n = len(arr)
res = [0] * n
def dp(i):
if res[i]:
return res[i]
res[i] = 1
for di in [-1, 1]:
for j in range(i + di, i + di + d * di, di):
if not (0 <= j < n and arr[j] < arr[i]):
break
res[i] = max(res[i], dp(j) + 1)
return res[i]
return max(map(dp, range(n))) | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR DICT FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF NUMBER BIN_OP VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR RETURN VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FUNC_DEF IF VAR VAR RETURN VAR VAR ASSIGN VAR VAR NUMBER FOR VAR LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR VAR IF NUMBER VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER RETURN VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, lst: List[int], d: int) -> int:
def Ab(i, d, X):
if i < 0 or i >= len(lst):
return -1
if X <= lst[i]:
return None
if dp[i] != -1:
return dp[i]
m = 1
for j in range(1, d + 1):
a = 0
a = Ab(i - j, d, lst[i])
if a == None:
break
m = max(m, 1 + a)
for j in range(1, d + 1):
b = 0
b = Ab(i + j, d, lst[i])
if b == None:
break
m = max(m, 1 + b)
dp[i] = m
return dp[i]
dp = [(-1) for i in lst]
for i in range(len(lst) - 1, -1, -1):
Ab(i, d, sys.maxsize)
return max(dp) | CLASS_DEF FUNC_DEF VAR VAR VAR FUNC_DEF IF VAR NUMBER VAR FUNC_CALL VAR VAR RETURN NUMBER IF VAR VAR VAR RETURN NONE IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR IF VAR NONE ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR IF VAR NONE ASSIGN VAR FUNC_CALL VAR VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR RETURN VAR VAR ASSIGN VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR |
Given an array of integers arr and an integer d. In one step you can jump from index i to index:
i + x where: i + x < arr.length and 0 < x <= d.
i - x where: i - x >= 0 and 0 < x <= d.
In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).
You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.
Example 2:
Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.
Example 3:
Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies.
Example 4:
Input: arr = [7,1,7,1,7,1], d = 2
Output: 2
Example 5:
Input: arr = [66], d = 1
Output: 1
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 10^5
1 <= d <= arr.length | class Solution:
def maxJumps(self, arr: List[int], d: int) -> int:
dp = [0] * len(arr)
path = {i: [(i, arr[i])] for i in range(len(arr))}
def jump(i):
nonlocal dp, path
if dp[i] != 0:
return dp[i]
left, right = 0, 0
nl, nr = i, i
for j in range(i + 1, i + d + 1):
if j >= len(arr):
break
if arr[j] < arr[i]:
t = jump(j)
if right < t:
right = t
nr = j
else:
break
for j in range(i - 1, i - d - 1, -1):
if j < 0:
break
if arr[j] < arr[i]:
t = jump(j)
if left < t:
left = t
nl = j
else:
break
if left < right and right > 0:
path[i] += path[nr]
elif left >= right and left > 0:
path[i] += path[nl]
dp[i] = max(left, right) + 1
return dp[i]
ans = 0
for i in range(len(arr)):
jump(i)
ans = max(ans, dp[i])
return ans | CLASS_DEF FUNC_DEF VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR LIST VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_DEF IF VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER IF VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER NUMBER IF VAR NUMBER IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR VAR IF VAR VAR VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR VAR |
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