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class Solution {
public:
int minSubArrayLen(int target, vector<int>& nums) {
int res=INT_MAX;
int l=0;
int r=0;
int sum=0;
while(r<nums.size()){
sum+=nums[r];
r++;
while(l<r && sum>=target){
sum-=nums[l];
l++;
res=min(res,r-l+1);
}
}
if(res==INT_MAX)
return 0;
return res;
}
};
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def smallestSubWithSum(arr, n, x):
# Initialize current sum and minimum length
curr_sum = 0
min_len = n + 1
# Initialize starting and ending indexes
start = 0
end = 0
while (end < n):
# Keep adding array elements while current
# sum is smaller than or equal to x
while (curr_sum <= x and end < n):
curr_sum += arr[end]
end += 1
# If current sum becomes greater than x.
while (curr_sum > x and start < n):
# Update minimum length if needed
if (end - start < min_len):
min_len = end - start
# remove starting elements
curr_sum -= arr[start]
start += 1
return min_len
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def maxSubArray(self, nums: List[int]) -> int:
curr_best = overall_best = nums[0]
for i in range(1,len(nums)):
if curr_best>=0:
curr_best = curr_best + nums[i]
else:
curr_best = nums[i]
if curr_best > overall_best:
overall_best = curr_best
return overall_best
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var maxSequence = function(arr){
var min = 0, ans = 0, i, sum = 0;
for (i = 0; i < arr.length; ++i) {
sum += arr[i];
min = Math.min(sum, min);
ans = Math.max(ans, sum - min);
}
return ans;
}
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l=[4,6,10,8,2,1]
ans=0
m=3
for i in range(len(l)+1):
for j in range(i):
if len(l[j:i])==m:
ans=max(ans,sum(l[j:i]))
print(ans)