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.

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).

Tips:

O(n): 两个指针, start end, end向后走,直到 sum 大于 s. 然后start向后, 直到sum 小于s. 同时更新 min值。类似于滑动窗口的形式。

O(nlogn): Since all elements are positive, the cumulative sum must be strictly increasing. Then, a subarray sum can expressed as the difference between two cumulative sum. Hence, given a start index for the cumulative sum array, the other end index can be searched using binary search.

Code:

O(n):

public class Solution {
    public int minSubArrayLen(int s, int[] a) {
      if (a == null || a.length == 0){
        return 0;
      }

      int i = 0, j = 0, sum = 0, min = Integer.MAX_VALUE;

      while (j < a.length) {
        sum += a[j++];

        while (sum >= s) {
          min = Math.min(min, j - i);
          sum -= a[i++];
        }
      }

      return min == Integer.MAX_VALUE ? 0 : min;
    }
}

O(nlogn):

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;
}