<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 {
const long long int MOD = 1000000007;
public:
    int threeSumMulti(vector<int>& arr, int target) {
        unordered_map<int, long long int> num_count;
        for(auto num: arr){
            num_count[num]++;
        }
        long long int count = 0;
        for(int num1=0;num1<101;num1++){
            for(int num2=num1;num2<101;num2++){
                int num3 = target-num2-num1;
                if(num1 == num2 && num2 == num3){
                    count += ((num_count[num1] * (num_count[num1]-1) * (num_count[num1]-2))/6);
                }else if(num1 == num2){
                    count += (((num_count[num1] * (num_count[num1]-1))/2)*num_count[num3]);
                }else if(num1 < num2 && num2 < num3){
                    count += (num_count[num1] * num_count[num2] * num_count[num3]);
                }
                count %= MOD;
            }
        }
        return count;
    }
};
/*
Think of cases:
target = 6
2 count 7
2 2 2
7 C 3 -> 7*6*5/3 -> 70
*/
15600