209. Minimum Size Subarray Sum.md

209. Minimum Size Subarray Sum

Difficulty : Medium

Related Topics : Array、Two Pointers、Binary Search

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leetcode-cn Daily Challenge on June 28, 2020.

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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).

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Solution

• mine
  • Java
    • Two Pointer Runtime: 1 ms, faster than 99.93%, Memory Usage: 39.6 MB, less than 43.11% of Java online submissions

      // O(n)time 
      // O(1)space
      public int minSubArrayLen(int s, int[] nums) {
          int len = nums.length;
          int res = len + 1;
          int pre = 0, cur = 0;
          int sum = 0;
          for(int i = 0; i < nums.length; i++){
              if(nums[i] >= s){
                  return 1;
              }
              sum += nums[i];
              cur++;
              while(sum >= s){
                  sum -= nums[pre];
                  res = Math.min(cur - pre, res);
                  pre++;
              }
          }
          return res == len + 1 ? 0 : res;
      }
      

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• the most votes
  • Two Pointer Runtime: 1 ms, faster than 99.93%, Memory Usage: 39.5 MB, less than 62.72% of Java online submissions

    // O(n)time 
    // O(1)space
    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;
    }
    

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