Problem statement:

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn‘t one, return 0 instead.

For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.

Solution:

It looks like 53. Maximum Subarray. But they are different solutions. I solve this problem by two pointers, left and right. The return value is the min length which is greater than or equal to the target. I can control the movement of these two pointers to update the min length.

Basic idea:

• If sum >= target, left moves to the end and minus the value of left, meanwhile update the min length.
• One terminating condition is left > right, that means we already find the min length is 1 and return. Otherwise, I find until the end.
• If sum < target, right moves to the end and plus the value of right.

Since left and right move at most n steps. Time complexity is O(2 * n) ---> O(n).

class Solution {public:    int minSubArrayLen(int s, vector<int>& nums) {        if(nums.empty()){            return 0;        }        int left = 0;         int right = 0;        int sum = nums[left];        int len = INT_MAX;        while(left <= right && left < nums.size() && right < nums.size()){            // if current sum is a solution            if(sum >= s){                // update the len                len = min(len, right - left + 1);                // update the sum                // minus first and move left towards end                sum -= nums[left++];            } else {                // the sum is still less than target                // move right to the end and update the sum                sum += nums[++right];            }        }        // return 0 if no answer        return len == INT_MAX ? 0 : len;    }};

209. Minimum Size Subarray Sum