# 3 Sum With Multiplicity

Updated: Mar 24, 2021

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.

**Constraints:**

3 <= arr.length <= 3000

0 <= arr[i] <= 100

0 <= target <= 300

**Solution:**

```
class Solution {
public int threeSumMulti(int[] arr, int target) {
Map<Integer, Long> frequency = new HashMap<>();
for(int i: arr){
long freq = frequency.getOrDefault(i, 0l);
frequency.put(i, freq+1l);
}
long ans =0;
for(Integer x: frequency.keySet()){
for(Integer y : frequency.keySet()){
int z = target -x -y;
if(frequency.containsKey(z)){
long xfreq = frequency.get(x);
long yfreq = frequency.get(y);
long zfreq = frequency.get(z);
if(x == y && x ==z){
ans += ((xfreq) * (xfreq-1)*(xfreq-2))/6;
} else if(x ==y && x!=z){
ans += ((xfreq)*(xfreq-1))/2 *zfreq;
} else if(x< y && y<z){
ans += ((xfreq*yfreq *zfreq));
}
}
ans = ans % 1000000007;
}
}
return (int) ans;
}
}
```