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 <= 400001 <= 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;
}
}