<p>Given a list of <code>dominoes</code>, <code>dominoes[i] = [a, b]</code> is <strong>equivalent to</strong> <code>dominoes[j] = [c, d]</code> if and only if either (<code>a == c</code> and <code>b == d</code>), or (<code>a == d</code> and <code>b == c</code>) - that is, one domino can be rotated to be equal to another domino.</p>

<p>Return <em>the number of pairs </em><code>(i, j)</code><em> for which </em><code>0 &lt;= i &lt; j &lt; dominoes.length</code><em>, and </em><code>dominoes[i]</code><em> is <strong>equivalent to</strong> </em><code>dominoes[j]</code>.</p>

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

<pre>
<strong>Input:</strong> dominoes = [[1,2],[2,1],[3,4],[5,6]]
<strong>Output:</strong> 1
</pre>

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

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

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

<ul>
	<li><code>1 &lt;= dominoes.length &lt;= 4 * 10<sup>4</sup></code></li>
	<li><code>dominoes[i].length == 2</code></li>
	<li><code>1 &lt;= dominoes[i][j] &lt;= 9</code></li>
</ul>

class Solution:
    def numEquivDominoPairs(self, dominoes: List[List[int]]) -> int:
        equivalentPairs = 0
        for i in dominoes:
            i.sort()
        while len(dominoes) > 0:
            pair = dominoes[0]
            pairCount = dominoes.count(pair)
            for i in range(pairCount):
                dominoes.remove(pair)
            equivalentPairs += 0.5 * (pairCount ** 2) + (-0.5 * pairCount)
        return int(equivalentPairs)
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