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 * 104dominoes[i].length == 21 <= dominoes[i][j] <= 9class Solution {
public int numEquivDominoPairs(int[][] dominoes) {
int[] freq = new int[100];
int count = 0;
for (int[] pair : dominoes) {
int pairSum = Math.min(10 * pair[0] + pair[1], 10 * pair[1] + pair[0]);
count += freq[pairSum];
freq[pairSum]++;
}
return count;
}
}