<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:
    int threeSumMulti(vector<int>& arr, int target) {
        sort(arr.begin(),arr.end()); 
        int ans=0;int mod=1e9+7;
        
        for(int i=0;i<arr.size()-2;i++){
            int sum=target-arr[i];
            int j=i+1,k=arr.size()-1;
            while(j<k){
                if(arr[j]+arr[k]==sum){
                    
                    int col=1,coz=1;
                    while(j<k && arr[j]==arr[j+1])
                    { j++;col++;}
                     
                    while(j<k && arr[k]==arr[k-1])
                    {k--;coz++;}
                    if(j!=k)
                     ans+=(col*coz)%mod;
                    if(j==k){
                        ans=(ans+col*(col-1)/2)%mod;
                    }
                    k--;j++;
                }
                else if(arr[j]+arr[k]<sum){
                    j++;
                }
                else{
                    k--;
                }
                
            }
            
        }
       
        return ans%mod;
        
    }
};
13000