Question

923. 3Sum With Multiplicity

中文文档

Description

Given an integer array arr, and an integer target, return the number of tuples i, j, k such that i < j < k and arr[i] + arr[j] + arr[k] == target.

As the answer can be very large, return it modulo 109 + 7.

 

Example 1:

Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (arr[i], arr[j], arr[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.

Example 2:

Input: arr = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.

Example 3:

Input: arr = [2,1,3], target = 6
Output: 1
Explanation: (1, 2, 3) occured one time in the array so we return 1.

 

Constraints:

• 3 <= arr.length <= 3000
• 0 <= arr[i] <= 100
• 0 <= target <= 300

Solutions

Solution 1: Enumeration

First, we use an array cnt of length $101$ to record the number of occurrences of each element in the array arr.

Then, we enumerate the elements in the array arr as the second element $b$ of the triplet, and first subtract one $b$ from cnt. Then, starting from the beginning of the array arr, we enumerate the first element $a$, and the third element $c$ is $target - a - b$. If the value of $c$ is within the range of the array cnt, then the answer is increased by $cnt[c]$.

Note the modulo operation of the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(C)$. Where $n$ is the length of the array arr; and $C$ is the length of the array cnt, in this problem $C = 101$.

class Solution:
  def threeSumMulti(self, arr: List[int], target: int) -> int:
    cnt = Counter(arr)
    ans = 0
    mod = 10**9 + 7
    for j, b in enumerate(arr):
      cnt[b] -= 1
      for i in range(j):
        a = arr[i]
        c = target - a - b
        ans = (ans + cnt[c]) % mod
    return ans

class Solution {
  private static final int MOD = (int) 1e9 + 7;

  public int threeSumMulti(int[] arr, int target) {
    int[] cnt = new int[101];
    for (int v : arr) {
      ++cnt[v];
    }
    long ans = 0;
    for (int j = 0; j < arr.length; ++j) {
      int b = arr[j];
      --cnt[b];
      for (int i = 0; i < j; ++i) {
        int a = arr[i];
        int c = target - a - b;
        if (c >= 0 && c <= 100) {
          ans = (ans + cnt[c]) % MOD;
        }
      }
    }
    return (int) ans;
  }
}

class Solution {
public:
  const int mod = 1e9 + 7;

  int threeSumMulti(vector<int>& arr, int target) {
    int cnt[101] = {0};
    for (int& v : arr) {
      ++cnt[v];
    }
    long ans = 0;
    for (int j = 0; j < arr.size(); ++j) {
      int b = arr[j];
      --cnt[b];
      for (int i = 0; i < j; ++i) {
        int a = arr[i];
        int c = target - a - b;
        if (c >= 0 && c <= 100) {
          ans += cnt[c];
          ans %= mod;
        }
      }
    }
    return ans;
  }
};

func threeSumMulti(arr []int, target int) int {
  const mod int = 1e9 + 7
  cnt := [101]int{}
  for _, v := range arr {
    cnt[v]++
  }
  ans := 0
  for j, b := range arr {
    cnt[b]--
    for i := 0; i < j; i++ {
      a := arr[i]
      c := target - a - b
      if c >= 0 && c <= 100 {
        ans += cnt[c]
        ans %= mod
      }
    }
  }
  return ans
}