# My C++ moving window solution, O(n) time, O(1) space; another binary search version added(O(nlogn) time)

1. O(n) time, O(1) space moving window method
using a moving window [start,end] to calculate the sum, first move end forward to get a sub-array with sum>=s, then move start forward to make sum < s, then move end again,..., until end reach the end of array.
``````class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int start=0, end=0, sum=0, len = nums.size(), res;
while(end<len)
{
while(sum<s && end<len) sum += nums[end++];
while(sum>=s) sum -=nums[start++];
res = min(res, end-start + 1);
}
return res>len?0:res;
}
};
``````
1. O(nlogn) time, O(n) space version, binary search method
``````class Solution {
public:
int binary_search(vector<int>&sum, int left, int target)
{
int right = sum.size()-1, mid;
while(left<=right)
{
mid = (left+right)>>1;
if(sum[mid]>=target) right = mid-1;
else left = mid+1;
}
return (left<sum.size()?left:-1);
}

int minSubArrayLen(int s, vector<int>& nums) {
int len = nums.size();
vector<int> sum(len+1,0);
int i, start = 0, end;
int res = len+1;

if(len>0)
{
for(i=1;i<=len;i++) sum[i] =sum[i-1] + nums[i-1];
while(start<len && (end = binary_search(sum, start+1, s+sum[start]))>=0 )
{
res = min(res, end-start);
start++;
}
}
return res>len?0:res;``````

• For the case [1,2] and s=4, the o(n) method doesn't work.

while(sum<s && end<len) sum +=nums[++end];//It will be out of range.

• Thanks for pointing this out and it is fixed in the updated version. Appreciate your comments.

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