pg

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.

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

The idea is clear and simple:

• Adding new numbers from right side.
• When the current sum is larger than s, delete numbers from left side and update the result at the same time.

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int minSubArrayLen(int s, vector<int>& nums) {
        int i = 0;
        int j = 0;
        int sum = 0;
        int res = INT_MAX;
        while(j < nums.size()){
            sum += nums[j++];
            while(sum >= s){
                res = min(res, j-i);
                sum -= nums[i++];
            }
        }
        return res == INT_MAX ? 0 : res;
    }