Question

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

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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

Solution 1

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

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;
  }
};

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
}