1128. Number-Of-Equivalent-Domino-Pairs

Easy question is light. The comments do the justification here.

We first count the number of equivalent dominoes -> for each count of equivalent dominoes, the number of possible pairs is equal to (count * (count - 1)) // 2.

We calculate this for all the count of equivalent dominoes and return the result.

class Solution(object):
    def numEquivDominoPairs(self, dominoes):
        """
        :type dominoes: List[List[int]]
        :rtype: int
        """
        # dominoes can only be from 1 - 9
        # we can identify equivalent dominoes by sorting and making a string
        # count this number
        # for each domino, we can just add the count sum of count - 1?
        # which is count(count - 1) / 2

        # if we have 3 of one
        # we can do 
        # 1 and 2, 1 and 3, 2 and 3 (3 pairs)
        # if we have 4
        # 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3 and 4  (6 pairs)
        count_map = {}
        count = 0

        for domino in dominoes:
            key = str(min(domino)) + str(max(domino))
            count_map[key] = count_map.get(key, 0) + 1

        for value in count_map.values():
            count += ((value * (value - 1)) // 2)

        return count

Time Complexity: O(n)

Space Complexity: O(n)

Time Taken: 6m 50s