<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>


#include <vector>
#include <algorithm>
#include <unordered_map>



class Solution {
public:
    int threeSumMulti(std::vector<int>& arr, int target) {
		long long result = 0;
		std::unordered_map<int, int> map;

		std::sort(arr.begin(), arr.end());

		for (int num : arr)
			++map[num];

		for (int i = 0; i < arr.size() - 2; ++i) {
			if (i != 0 && arr[i] == arr[i - 1])
				continue;
			if (arr[i] > target)
				break;

			//	first == second == third => C(first_count, 3)
			if (target == arr[i] * 3)
				result += (long long)map[arr[i]] * (map[arr[i]] - 1) * (map[arr[i]] - 2) / (3 * 2 * 1);

			//	first == second != third => C(first_count, 2) * third_count
			if (arr[i] < target - arr[i] * 2 && map.count(target - arr[i] * 2))
				result += (long long)map[arr[i]] * (map[arr[i]] - 1) / (2 * 1) * map[target - arr[i] * 2];

			for (int j = i + 1; j < arr.size() - 1; ++j) {
				if (arr[j] == arr[j - 1])
					continue;
				if (arr[i] + arr[j] > target)
					break;

				//	first != second == third => first_count * C(second_count, 2)
				if (target == arr[i] + arr[j] * 2)
					result += (long long)map[arr[i]] * map[arr[j]] * (map[arr[j]] - 1) / (2 * 1);

				//	first != second != third => first_count * second_count * third_count
				if (arr[j] < target - arr[i] - arr[j] && map.count(target - arr[i] - arr[j]))
					result += (long long)map[arr[i]] * map[arr[j]] * map[target - arr[i] - arr[j]];
			}
		}

		return result % 1000000007;
    }
};
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