<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>

void print(vector<vector<int>>& mat) {
    for (auto v : mat) {
        for (auto i : v) {
            cout << i << ' ';
        }
        cout << '\n';
    }
    cout << '\n';
}

class Solution {
    using Mat = vector<vector<int>>;
public:
    int numberOfBoomerangs(Mat& points) {
        const int n = points.size();
    

        vector<map<int, int>> mem(n);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                const int x = points[i][0] - points[j][0];
                const int y = points[i][1] - points[j][1];
                const int dist = x*x + y*y;
                
                ++mem[i][dist];
            }
        }

        return accumulate(mem.begin(), mem.end(), 0, [n](int res, map<int, int>& m) {
            for (auto [dist, n] : m) {
                if (n > 1) {
                    res += n * (n-1);
                }
            }
            return res;
        });
    }
};

149718