# 923. 3Sum With Multiplicity

## 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

• 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;
}
};

• 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


• 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
}

• function threeSumMulti(arr: number[], target: number): number {
const mod = 10 ** 9 + 7;
const cnt: number[] = Array(101).fill(0);
for (const x of arr) {
++cnt[x];
}
let ans = 0;
const n = arr.length;
for (let j = 0; j < n; ++j) {
--cnt[arr[j]];
for (let i = 0; i < j; ++i) {
const c = target - arr[i] - arr[j];
if (c >= 0 && c < cnt.length) {
ans = (ans + cnt[c]) % mod;
}
}
}
return ans;
}