<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>& nums, int target) {
    //     int size = nums.size();
    //     sort(nums.begin(), nums.end());
    //     int count = 0;
    //     const int MOD = 1e9 + 7;
    //     for(int i=0; i<size-2; i++)
    //     {
    //         int l = i+1;
    //         int r = size - 1;
    //         while(l < r)
    //         {
    //             if((nums[i] + nums[l] + nums[r]) == target)
    //             {
    //                 count = count%MOD + 1;
    //             }
    //             if(r - l == 1)
    //             {
    //                 l++;
    //                 r = size - 1;
    //             }
    //             else r--;
    //         }
    //     }
    //     return count;
    // }
     int threeSumMulti(vector<int>& nums, int target) {
        
        sort(nums.begin(),nums.end());
        
        // two pointer approach 
		
        long long mod=1e9+7;
		
        int n=nums.size();
		
        int ans=0;
		
        long long sum;
		
        for(int i=0;i<n-2;i++)
        {
            int l=i+1;
			
            int r=n-1;
            
            while(l<r)
            {
                sum=nums[l]+nums[i]+nums[r];
             
                if(sum==target)
                {
                      int count1=1;
                      int count2=1;
                    while(l<r && nums[l]==nums[l+1])
                    {
                        l++;
                        count1++;
                    }
                    
                    while(l<r && nums[r]==nums[r-1])
                    {
                        r--;
                        count2++;
                    }
                                           
                    if(l==r)         
                    {                                  
                        ans = (ans%mod + (count1%mod *(count1-1)%mod)/2)%mod;
                    }
                    
                    if(l!=r)
                    {
                        ans = (ans%mod + (count1%mod* count2%mod)%mod)%mod;	 
                    }
                    
                    l++;
                    r--;
                }
                else
                {
                    if(sum>target)
                    {
                        r--;
                    }
                    else
                    {
                        l++;
                    }
                }
            }
        }
        
        return ans;
        
    }
};
13300