Problem Statement:

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

Solution:

 
class Solution {
public:
    int threeSumMulti(vector<int>& arr, int target) {
        int n = arr.size(), MAX=101, res=0, MOD=1000000007;
        vector<int> frequencies(MAX, 0);
        for (int k: arr) frequencies[k]++;
        for (int x=0; x<MAX; x++)
        {
            for (int y=x+1; y<MAX; y++)
            {
                int z=target-x-y;
                if (z>y && z<MAX)
                {
                    long long a=frequencies[x], b=frequencies[y], c=frequencies[z];
                    long long d = a*b*c;
                    res += d%MOD;
                    res %= MOD;
                }
            }
        }
        for (int x=0; x<MAX; x++)
        {
            int y = target-2*x; //x,x,y
            if (y>x && y<MAX)
            {
                long long a=frequencies[x], b=frequencies[y];
                long long c = b * a * (a-1)/2;
                res += c%MOD;
                res %= MOD;
            }
        }
        for (int x=0; x<MAX; x++)
        {
            int y = (target-x)/2; // x,y,y
            if (y>x && (target-x)%2==0 && y<MAX)
            {
                long long a=frequencies[x], b=frequencies[y];
                long long c = a * b * (b-1)/2;
                res += c%MOD;
                res %= MOD;
            }
        }
        if (target%3==0)
        {
            int x=target/3;
            long long a=frequencies[x];
            long long b = a * (a-1) * (a-2)/6;
            res += b%MOD;
            res %= MOD;
        }
        return res;
    }
};

 
TC: O(MAX^2)
SC: O(MAX)

where MAX is the maximum value in the array (here 100). Notice that the complexities do not depend on the length of arr.