Description

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.

click to show more practice.

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).

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

Solution

Two-pointer, time O(n), space O(1)

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int minLength = Integer.MAX_VALUE;
        int i = 0;
        int sum = 0;
        
        for (int j = 0; j < nums.length; ++j) {
            sum += nums[j];
            if (sum < s) {
                continue;
            }
            
            while (sum - nums[i] >= s) {
                sum -= nums[i++];
            }
            
            minLength = Math.min(j - i + 1, minLength);
        }
        
        return minLength == Integer.MAX_VALUE ? 0 : minLength;
    }
}

Binary-search, time O(nlogn), space O(n)

有趣的解法。

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        return solveNLogN(s, nums);
    }

    private int solveNLogN(int s, int[] nums) {
        int[] sums = new int[nums.length + 1];
        for (int i = 1; i < sums.length; i++) sums[i] = sums[i - 1] + nums[i - 1];
        int minLen = Integer.MAX_VALUE;
        for (int i = 0; i < sums.length; i++) {
            int end = binarySearch(i + 1, sums.length - 1, sums[i] + s, sums);
            if (end == sums.length) break;
            if (end - i < minLen) minLen = end - i;
        }
        return minLen == Integer.MAX_VALUE ? 0 : minLen;
    }
    
    private int binarySearch(int lo, int hi, int key, int[] sums) {
        while (lo <= hi) {
           int mid = (lo + hi) / 2;
           if (sums[mid] >= key){
               hi = mid - 1;
           } else {
               lo = mid + 1;
           }
        }
        return lo;
    }
}