<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:

    const int mod = 1e9+7;

    int threeSumMulti(vector<int>& arr, int target) {
        int n = arr.size();
        sort(arr.begin(),arr.end());
        map<int,int> mp;
        set<vector<int> > s;
        for(auto c : arr) mp[c]++;
        long long ans = 0;
        for(int i=0;i<n;i++)
        {
            int j = i+1 , k = n-1;
            while(j<k)
            {
                if(arr[i]+arr[j]+arr[k]<target)
                {
                    j++;
                }
                else if(arr[i]+arr[j]+arr[k]>target)
                {
                    k--;
                }
                else
                {
                    int x = arr[i];
                    int y = arr[j];
                    int z = arr[k];

                   s.insert({x,y,z});

                    j++;
                    k--;
                    while(j<n && arr[j]==arr[j-1]) j++;
                    while(k>=0 && arr[k]==arr[k+1]) k--;
                }
            }
        }

        while(!s.empty())
        {
            auto v = *s.begin();
            s.erase(s.begin());
            int x = v[0];
            int y = v[1];
            int z = v[2];
            if(x!=y && y!=z)
            {
                ans+=((1LL*mp[x]*1LL*mp[y])%mod*1LL*mp[z])%mod;
            }
            else if(x==y && y!=z or x!=y && y==z)
            {
                if(x==y)
                {
                    ans+=((((1LL*mp[x])*(1LL*mp[x]-1)/2)%mod)*1LL*mp[z])%mod;
                }
                else if(y==z)
                {
                    ans+=((((1LL*mp[y])*(1LL*mp[y]-1)/2)%mod)*1LL*mp[x])%mod;
                }
            }
            else
            {
                ans+=(((((1LL*mp[x])*(1LL*mp[x]-1))%mod)*(1LL*mp[x]-2))/6)%mod;;
            }

        }
        return ans;
    }
};
27300