1128. Number of Equivalent Domino Pairs Easy

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

Approach

Input: An array dominoes

Output: Determine how many pairs of equivalent dominoes there are

This is a Hash Table + Array Counting problem.

We can sort the numbers on each domino and convert them into a string or tuple to store in a hash table.

When iterating to a new domino, check the hash table for how many equivalent dominoes have appeared before, and increment the count: count += cache[key].

Finally, return the answer count.

Implementation

python

class Solution:
    def numEquivDominoPairs(self, dominoes: List[List[int]]) -> int:
        cache = {}  # Used to count occurrences of each equivalent pair
        count = 0   # Final result: number of equivalent pairs

        for a, b in dominoes:
            # Sort the two numbers on each domino in ascending order, ensuring [1,2] and [2,1] fall into the same category
            key = (min(a, b), max(a, b))

            # If this combination has appeared before, there are already count[key] dominos that can be paired with the current one
            if key in cache:
                count += cache[key]  # This current domino can form a pair with each of the existing equivalent dominos
                cache[key] += 1      # Update the occurrence count
            else:
                cache[key] = 1       # First appearance, initialized to 1

        return count

javascript

/**
 * @param {number[][]} dominoes
 * @return {number}
 */
var numEquivDominoPairs = function(dominoes) {
    const cache = new Map();
    let count = 0;
    for (let arr of dominoes) {
        arr = [Math.min(...arr), Math.max(...arr)];
        const key = String(arr);

        if (cache.has(key)) {
            count += cache.get(key);
            cache.set(key, cache.get(key) + 1);
        } else {
            cache.set(key, 1);
        }
    }

    return count;
};

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(1) space since there are at most 100 possible domino pairs (from [1,1] to [9,9])

Links

1128. Number of Equivalent Domino Pairs (English)1128. 等价多米诺骨牌对的数量 (Chinese)