1128. Number of Equivalent Domino Pairs

#Problem

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

Return the number of pairs (i, j) for which 0 <= i < j < dominoes.length, and dominoes[i] is equivalent to dominoes[j].

https://leetcode.com/problems/number-of-equivalent-domino-pairs/

Example 1:

Input: dominoes = [[1,2],[2,1],[3,4],[5,6]]
Output: 1

Example 2:

Input: dominoes = [[1,2],[1,2],[1,1],[1,2],[2,2]]
Output: 3

Constraints:

• 1 <= dominoes.length <= 4 * 10⁴
• dominoes[i].length == 2
• 1 <= dominoes[i][j] <= 9

#Test Cases

class Solution:
    def numEquivDominoPairs(self, dominoes: List[List[int]]) -> int:

import pytest

from solution import Solution


@pytest.mark.parametrize('dominoes, expected', [
    ([[1,2],[2,1],[3,4],[5,6]], 1),
    ([[1,2],[1,2],[1,1],[1,2],[2,2]], 3),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, dominoes, expected):
    assert sol.numEquivDominoPairs(dominoes) == expected

#Thoughts

把所有的多米诺骨牌都翻转成小面在前（即 a ≤ b），然后统计各种牌的数量。如果某种牌有 k 个，当 k > 1 时，其中任意两个都是一对，共有 k * (k - 1) / 2 对。

累加每种牌的对数即可。

时间复杂度 O(n)，空间复杂度 O(n)。

#Code

from typing import Counter


class Solution:
    def numEquivDominoPairs(self, dominoes: list[list[int]]) -> int:
        counts = Counter((a, b) if a <= b else (b, a) for a, b in dominoes)
        return sum(k * (k-1) // 2 for k in counts.values() if k > 1)