<p>You are given <code>n</code> <code>points</code> in the plane that are all <strong>distinct</strong>, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>. A <strong>boomerang</strong> is a tuple of points <code>(i, j, k)</code> such that the distance between <code>i</code> and <code>j</code> equals the distance between <code>i</code> and <code>k</code> <strong>(the order of the tuple matters)</strong>.</p>

<p>Return <em>the number of boomerangs</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> points = [[0,0],[1,0],[2,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> points = [[1,1],[2,2],[3,3]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> points = [[1,1]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == points.length</code></li>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>points[i].length == 2</code></li>
	<li><code>-10<sup>4</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li>All the points are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    int numberOfBoomerangs(vector<vector<int>>& points) {

        if (points.size()==1)
            return 0;

        vector<vector<double>> distances(points.size(), vector<double> (points.size(), 0));

        for (int i{}; i<points.size(); i++) {
            for (int j {i+1}; j<points.size(); j++) {
                distances[i][j] = distance(points[i], points[j]);
                distances[j][i] = distances[i][j];
            };
        };

        int i = 0, result = 0;
        for (vector<double> row : distances) {

            row.erase(row.begin() + i);
            i++;

            unordered_map<double, int> counts{};

            for (const auto& ele : row) {
                if (counts.find(ele)==counts.end())
                    counts[ele] = 1;
                else 
                    counts[ele]++;
            };

            for (const auto& [val, count] : counts) {
                if (count>1)
                    result += nPk(count);
            };
        };

        return result;
    };

    double distance(const vector<int>& a, const vector<int>& b) {
        return sqrt((a[0]-b[0])*(a[0]-b[0]) + (a[1]-b[1])*(a[1]-b[1]));
    };

    int nPk(int n) {

        if (n==2)
            return 2;

        int result = 1, limit = n-2;
        while (n>limit) 
            result *= n--;
        return result;
    };
};
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