<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) {
        int MOD = 1000000007;
        long result = 0;
        long cache[101] = { 0 };
        for (int i = 0; i < arr.size(); ++i) {
            ++cache[arr[i]];
        }

        for (int i = 0; i <= 100; ++i) {
            if (cache[i] == 0) {
                continue;
            }
            for (int j = 0; j <= 100; ++j) {
                if (cache[j] == 0) {
                    continue;
                }
                int k = target - i - j;
                if (k < 0 || k > 100 || cache[k] == 0) {
                    continue;
                }
                if (i == k && j == k) {
                    result += (cache[i] * (cache[i] - 1) * (cache[i] - 2) / 6);
                } else if (i == j && j != k) {
                    result += ((cache[i] * (cache[i] - 1)) / 2 * cache[k]);
                } else if (i < j && j < k) {
                    result += (cache[i] * cache[j] * cache[k]);
                }
            }
        }
        return result % MOD;
    }
};
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