<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>& v, int t) {

        long long i, j, k, n = v.size(), cnt = 0;

        long long m = 1e9 + 7;

        map<long long, long long> mp;

        for (i = 0; i < n; i++) {
            mp[v[i]]++;
        }

        long long x, y, z;

        if (t == 0) {
            x = mp[0];
           
            y= x*(x - 1)*(x - 2) / 6;

            cout<<y<<endl;

            if (y > 0 && x > 0) {
                // cout<<"iii   "<<i<<"   "<<y<<endl;
                cnt = (cnt % m + y % m) % m;
            }

            return cnt;
        }

        //cout<<"yes3"<<endl;


        for (i = 0; i <= t; i++) {
            k = t - (i + i);

            if (k != i) {

                x = mp[i] - 1;
                y = (x * (x + 1)) / 2;
                z = y * mp[k];
                // cout<<"yes3"<<endl;
                if (z > 0)
                    cnt = (cnt % m + z % m) % m;
            }

            if (i + i + i == t) {
                  x = mp[i];
           
                  y= x*(x - 1)*(x - 2) / 6;

                if (y > 0 && x > 0) {
                    // cout<<"iii   "<<i<<"   "<<y<<endl;
                    cnt = (cnt % m + y % m) % m;
                }
            }

            for (j = i + 1; j <= t; j++) {
                k = t - (i + j);
                if (k > j) {
                    // cout<<"yes"<<endl;
                    //  cout<<i<<"  "<<j<<"  "<<k<<endl;
                    //  cout<<mp[i]<<"  "<<mp[j]<<"  "<<mp[k]<<endl;

                    cnt =
                        (cnt % m + (mp[i] % m * mp[j] % m * mp[k] % m) % m) % m;
                }
            }
        }

        return cnt;
    }
};
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