#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 <= 10^51 <= arr[i] <= 10001 <= target <= 10^8
class Solution {
public:
int minSumOfLengths(vector& 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 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;
}
};