209. Minimum Size Subarray Sum

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.

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).

Solution:

Suppose we know the prefixSum of each index i.

At any index i, we check if it is >= target. If so, we decrease its range from the left side, and check if it still satisfy the condition.

We can do this in O(n) time and without using extra space.



Code:

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