<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:
    def numberOfBoomerangs(self, points: List[List[int]]) -> int:
        # {pointI: {distrance: count}}
        dists = {}
        for p1 in points:
            if (p1[0], p1[1]) not in dists:
                dists[(p1[0], p1[1])] = {}
            for p2 in points:
                if p1 == p2:
                    continue
                dist = (p1[0]-p2[0])**2 + (p1[1] - p2[1])**2
                if dist not in dists[(p1[0], p1[1])]:
                    dists[(p1[0], p1[1])][dist] = 0
                dists[(p1[0], p1[1])][dist] += 1
        rest = 0
        for p in dists:
            for dist in dists[p]:
                cnt = dists[p][dist]
                rest += cnt * (cnt - 1)
        return rest
                
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