README_EN.md

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

Python3

Java

C++

class Solution {
public:
    int threeSumMulti(vector<int>& arr, int target) {
        unordered_map<int, long> c;
        for (int a : arr) c[a]++;
        long res = 0;
        for (auto it : c)
            for (auto it2 : c) {
                int i = it.first, j = it2.first, k = target - i - j;
                if (!c.count(k)) continue;
                if (i == j && j == k)
                    res += c[i] * (c[i] - 1) * (c[i] - 2) / 6;
                else if (i == j && j != k)
                    res += c[i] * (c[i] - 1) / 2 * c[k];
                else if (i < j && j < k)
                    res += c[i] * c[j] * c[k];
            }
        return res % int(1e9 + 7);
    }
};