<p>Given an integer array <code>arr</code>, and an integer <code>target</code>, return the number of tuples <code>i, j, k</code> such that <code>i &lt; j &lt; k</code> and <code>arr[i] + arr[j] + arr[k] == target</code>.</p>

<p>As the answer can be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,1,2,2,3,3,4,4,5,5], target = 8
<strong>Output:</strong> 20
<strong>Explanation: </strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,1,2,2,2,2], target = 5
<strong>Output:</strong> 12
<strong>Explanation: </strong>
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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [2,1,3], target = 6
<strong>Output:</strong> 1
<strong>Explanation:</strong> (1, 2, 3) occured one time in the array so we return 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= arr.length &lt;= 3000</code></li>
	<li><code>0 &lt;= arr[i] &lt;= 100</code></li>
	<li><code>0 &lt;= target &lt;= 300</code></li>
</ul>

class Solution {
public:
    long long comb(int total, int pick){
        const int MOD = 1e9 + 7;
        if(total < pick) return 0;
        if(total == pick) return 1;
        long long res = 1, divide = 1;
        for(int i = 1; i <= pick; i++){
            res *= (total - i + 1);
            divide *= i;
        }
        return (res / divide) % MOD;
    }

    int threeSumMulti(vector<int>& arr, int target) {
        const int MOD = 1e9 + 7;
        map<int, int> table;
        for(auto& num : arr){
            table[num]++;
        }

        vector<int> sorted_arr;
        for(auto& [num, freq] : table){
            sorted_arr.push_back(num);
        }

        long res = 0;
        for(int i = 0; i < sorted_arr.size(); i++){
            int l = i, r = sorted_arr.size() - 1;
            while(l <= r){
                int sum = sorted_arr[i] + sorted_arr[l] + sorted_arr[r];
                if(sum == target){
                    map<int, int> tmp;
                    long tmpRes = 1;
                    tmp[sorted_arr[i]]++;
                    tmp[sorted_arr[l]]++;
                    tmp[sorted_arr[r]]++;
                    for(auto [num, freq] : tmp){
                        tmpRes *= comb(table[num], freq);
                    }
                    res = (res + tmpRes) % MOD;
                    l++;
                    r--;
                } else if(sum < target){
                    l++;
                } else {
                    r--;
                }
            }
        }
        return res;
    }
};

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