README.md

923. 3Sum With Multiplicity

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 (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn three_sum_multi(arr: Vec<i32>, target: i32) -> i32 {
        let mut hashmap = HashMap::new();
        let mut ret = 0;

        for j in 0..arr.len() {
            ret = (ret + hashmap.get(&(target - arr[j])).unwrap_or(&0)) % 1_000_000_007;

            for i in 0..j {
                hashmap
                    .entry(arr[i] + arr[j])
                    .and_modify(|c| *c += 1)
                    .or_insert(1);
            }
        }

        ret
    }
}