1128. Number of Equivalent Domino Pairs

Given a list of dominoes, dominoes[i] = [a, b] is equivalentto 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

Constraints:

• 1 <= dominoes.length <= 40000
• 1 <= dominoes[i][j] <= 9

hash table (naive): 把一个domino pair存入set，再作为key存入map，遍历dominoes，计数每个set出现多少次，最后再计算组合数求pair num

time: O(n), space: O(n)

class Solution {
    public int numEquivDominoPairs(int[][] dominoes) {
        Map<Set<Integer>, Integer> map = new HashMap<>();
        for(int[] domino : dominoes) {
            Set<Integer> set = new HashSet<>();
            set.add(domino[0]);
            set.add(domino[1]);
            
            map.put(set, map.getOrDefault(set, 0) + 1);
        }
        
        int res = 0;
        for(Map.Entry<Set<Integer>, Integer> entry: map.entrySet()) {
            int n = entry.getValue();
            if(n >= 2) {
                res += n * (n - 1) / 2;
            }
        }
        return res;
    }
}

optimized: 把一个domino pair处理成一个两位的数字，再作为key存入map

time: O(n), space: O(n)

class Solution {
    public int numEquivDominoPairs(int[][] dominoes) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int[] domino : dominoes) {
            int max = Math.max(domino[0], domino[1]);
            int min = Math.min(domino[0], domino[1]);
            int n = min * 10 + max;
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        
        int res = 0;
        for(Map.Entry<Integer, Integer> entry: map.entrySet()) {
            int n = entry.getValue();
            if(n >= 2) {
                res += n * (n - 1) / 2;
            }
        }
        return res;
    }
}