#1477. Find Two Non-overlapping Sub-arrays Each With Target Sum

题目描述：

Given an array of integers arr and an integer target.

You have to find two non-overlapping sub-arrays of arr each with sum equal target. There can be multiple answers so you have to find an answer where the sum of the lengths of the two sub-arrays is minimum.

Return the minimum sum of the lengths of the two required sub-arrays, or return -1 if you cannot find such two sub-arrays.

Example 1:

Input: arr = [3,2,2,4,3], target = 3
Output: 2
Explanation: Only two sub-arrays have sum = 3 ([3] and [3]). The sum of their lengths is 2.

Example 2:

Input: arr = [7,3,4,7], target = 7
Output: 2
Explanation: Although we have three non-overlapping sub-arrays of sum = 7 ([7], [3,4] and [7]), but we will choose the first and third sub-arrays as the sum of their lengths is 2.

Example 3:

Input: arr = [4,3,2,6,2,3,4], target = 6
Output: -1
Explanation: We have only one sub-array of sum = 6.

Example 4:

Input: arr = [5,5,4,4,5], target = 3
Output: -1
Explanation: We cannot find a sub-array of sum = 3.

Example 5:

Input: arr = [3,1,1,1,5,1,2,1], target = 3
Output: 3
Explanation: Note that sub-arrays [1,2] and [2,1] cannot be an answer because they overlap.

Constraints:

• 1 <= arr.length <= 10^5
• 1 <= arr[i] <= 1000
• 1 <= target <= 10^8

class Solution {
public:
    int minSumOfLengths(vector<int>& arr, int target) {
        // 由于都是正数，一个起点最多对应一个终点，所有可能的子数组最多有n个
        // 也是由于都是正数才可以用滑动窗口
        int result = INT_MAX;
        const int n = arr.size();
        // min_length[i]表示在i和i之前的合法子数组的最小长度
        vector<int> min_lens(n, INT_MAX);
        int sum = 0;
        int i = 0, j = 0;
        int min_len = INT_MAX;
        while(j < n) {
            sum += arr[j];
            while(i <= j && sum > target) {
                sum -= arr[i];
                i++;
            }
            if (sum == target) {
                int cur_len = j - i + 1;
                min_len = min(min_len, cur_len);
                // 当发现当前合法子数组之前已经有了合法子数组，则更新结果
                if (i > 0 && min_lens[i - 1] != INT_MAX) {
                    result = min(result, cur_len + min_lens[i - 1]);
                }
            }
            // 最后再更新min_lens
            min_lens[j] = min_len;
            j++;
        }
        return result == INT_MAX ? -1 : result;
    }
};