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++;
}
if (r==nums.length && l==0 && sum<s) return 0;
return minWin;
}
}

Initially I'm concerning I should maintain the window to have its sum always be >= s, and l should always fall at a place maintaining this property. But the solution above proves that I need not do this. Anyway, my solution also works as follows:

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];
if (sum >= s) { //found one feasible window, try to shrink the window
while (l<=r && sum-nums[l] >= s) {
sum -= nums[l++];
}
minWin = Math.min(minWin, r-l+1);
}
r++;
}
if (r==nums.length && l==0 && sum<s) return 0;
return minWin;
}
}