Easy

Number of Equivalent Domino PairsPython

Full explanation · Time O(n) · Space O(n)

# Time:  O(n)
# Space: O(n)

import collections


class Solution(object):
    def numEquivDominoPairs(self, dominoes):
        """
        :type dominoes: List[List[int]]
        :rtype: int
        """
        counter = collections.Counter((min(x), max(x)) for x in dominoes)
        return sum(v*(v-1)//2 for v in counter.itervalues())