Number of Equivalent Domino Pairs

main.md

Question

Approach

To solve this, I realized that each domino can be uniquely represented by its sorted tuple form — for example, [2,1] becomes (1,2). This ensures that both [1,2] and [2,1] are considered equal. I then used a dictionary to count how many times each normalized domino appears. For each count greater than 1, I computed the number of unique pairs using the combination formula count * (count - 1) // 2, and summed these up to get the final answer.

Implementation

class Solution:
    def numEquivDominoPairs(self, dominoes: List[List[int]]) -> int:
        count = defaultdict(int)
        res = 0
        
        for a, b in dominoes:
            key = tuple(sorted((a, b)))
            res += count[key]
            count[key] += 1
            
        return res

Complexities

• Time: O(n)
• Space: O(n)