Leetcode: Minimum Size Subarray Sum

Given an array of n positive integers and a positive integer s, find the minimal length of a 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.

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More practice:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

Haven't think about the O(nlogn) solution.

O(n) solution is to maintain a window

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int minWin = Integer.MAX_VALUE;
        int l=0, r=0;
        if (nums==null || nums.length==0) return 0;
        int sum = 0;
        while (r < nums.length) {
            sum += nums[r];
            while (sum >= s) {
                minWin = Math.min(minWin, r-l+1);
                sum -= nums[l++];
            }
            r++;
        }
        return minWin == Integer.MAX_VALUE? 0 : minWin;
    }
}

My solution: shrink as much as possible while maintaining the window always correct

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int minWin = Integer.MAX_VALUE;
        int l=0, r=0;
        if (nums == null || nums.length == 0) return 0;
        int sum = 0;
        while (r < nums.length) {
            sum += nums[r];
            while (sum - nums[l] >= s) {
                sum -= nums[l++];
            }
            if (sum >= s) 
                minWin = Math.min(minWin, r - l + 1);
            r++;
        }
        
        return minWin == Integer.MAX_VALUE? 0 : minWin;
    }
}