MinimumSizeSubarrayWithSumGreaterThanK.java

package Leetcode;

/**
 * @author kalpak
 *
 * 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.
 *
 * Example:
 * 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.
 * Follow up:
 * If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
 */

public class MinimumSizeSubarrayWithSumGreaterThanK {
    public static int minSubArrayLen(int s, int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;

        int result = Integer.MAX_VALUE;
        int windowSum = 0;
        int i = 0;
        int j = 0;

        while(j < nums.length) {
            windowSum += nums[j];

            if(windowSum < s)
                j++;
            else if(windowSum >= s) {
                // Shrink the window
                // Important to note that we need to perform the check for minimum window inside the while loop for the best window and not outside, since we are interested in a lower bound.
                while(windowSum >= s) {
                    result = Math.min(result, j - i + 1);
                    windowSum -= nums[i];
                    i++;
                }
                j++;
            }
        }
        return result == Integer.MAX_VALUE ? 0 : result;
    }

    public static void main(String[] args) {
        int[] arr = new int[]{2,3,1,2,4,3};
        System.out.println(minSubArrayLen(7, arr));
    }
}