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

## Description

You are given an array of integers arr and an integer target.

You have to find two non-overlapping sub-arrays of arr each with a 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.

Constraints:

• 1 <= arr.length <= 105
• 1 <= arr[i] <= 1000
• 1 <= target <= 108

## Solutions

• class Solution {
public int minSumOfLengths(int[] arr, int target) {
Map<Integer, Integer> d = new HashMap<>();
d.put(0, 0);
int n = arr.length;
int[] f = new int[n + 1];
final int inf = 1 << 30;
f[0] = inf;
int s = 0, ans = inf;
for (int i = 1; i <= n; ++i) {
int v = arr[i - 1];
s += v;
f[i] = f[i - 1];
if (d.containsKey(s - target)) {
int j = d.get(s - target);
f[i] = Math.min(f[i], i - j);
ans = Math.min(ans, f[j] + i - j);
}
d.put(s, i);
}
return ans > n ? -1 : ans;
}
}

• class Solution {
public:
int minSumOfLengths(vector<int>& arr, int target) {
unordered_map<int, int> d;
d[0] = 0;
int s = 0, n = arr.size();
int f[n + 1];
const int inf = 1 << 30;
f[0] = inf;
int ans = inf;
for (int i = 1; i <= n; ++i) {
int v = arr[i - 1];
s += v;
f[i] = f[i - 1];
if (d.count(s - target)) {
int j = d[s - target];
f[i] = min(f[i], i - j);
ans = min(ans, f[j] + i - j);
}
d[s] = i;
}
return ans > n ? -1 : ans;
}
};

• class Solution:
def minSumOfLengths(self, arr: List[int], target: int) -> int:
d = {0: 0}
s, n = 0, len(arr)
f = [inf] * (n + 1)
ans = inf
for i, v in enumerate(arr, 1):
s += v
f[i] = f[i - 1]
if s - target in d:
j = d[s - target]
f[i] = min(f[i], i - j)
ans = min(ans, f[j] + i - j)
d[s] = i
return -1 if ans > n else ans

• func minSumOfLengths(arr []int, target int) int {
d := map[int]int{0: 0}
const inf = 1 << 30
s, n := 0, len(arr)
f := make([]int, n+1)
f[0] = inf
ans := inf
for i, v := range arr {
i++
f[i] = f[i-1]
s += v
if j, ok := d[s-target]; ok {
f[i] = min(f[i], i-j)
ans = min(ans, f[j]+i-j)
}
d[s] = i
}
if ans > n {
return -1
}
return ans
}