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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
}

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