1128. Number of Equivalent Domino Pairs

Description

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].

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

💡 Hint 1:
For each domino j, find the number of dominoes you've already seen (dominoes i with i < j) that are equivalent.

💡 Hint 2:
You can keep track of what you've seen using a hashmap.

Submitted Code

class Solution:
    def numEquivDominoPairs(self, dominoes: List[List[int]]) -> int:
        hashmap = {}
        result = 0

        for d in dominoes:
            key = tuple(sorted(d))                  # 정렬하여 통일 후 타입 변경
            hashmap[key] = hashmap.get(key, 0) + 1

        for count in hashmap.values():
            if count > 1:
                result += count * (count - 1) // 2  # 조합 C(n, 2)
        
        return result

Runtime: 3 ms | Beats 98.02%
Memory: 23.98 MB | Beats 91.35%

sorted()로 각 도미노의 방향을 통일해서 저장했다. 또 값을 변경 불가능한 immutable 타입만 딕셔너리의 키로 사용 가능하기 때문에 리스트를 튜플로 변경했다.

Other Solutions

1st

class Solution(object):
    def numEquivDominoPairs(self, dominoes):
        mpp = [0] * 100
        count = 0
        for a, b in dominoes:
            key = a * 10 + b if a <= b else b * 10 + a
            count += mpp[key]
            mpp[key] += 1
        return count

time complexity: 𝑂(𝑛)
space complexity: 𝑂(1)

정렬이나 딕셔너리 없이 고정된 크기를 가진 리스트를 사용하는 최적화된 버전이다.