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] <= 9
class Solution:
def numEquivDominoPairs(self, dominoes: List[List[int]]) -> int:
cnt = Counter()
ans = 0
for a, b in dominoes:
ans += cnt[(a, b)]
cnt[(a, b)] += 1
if a != b:
cnt[(b, a)] += 1
return ansclass Solution:
def numEquivDominoPairs(self, dominoes: List[List[int]]) -> int:
cnt = Counter()
ans = 0
for a, b in dominoes:
x = a * 10 + b if a < b else b * 10 + a
ans += cnt[x]
cnt[x] += 1
return ansclass Solution {
public int numEquivDominoPairs(int[][] dominoes) {
int[] cnt = new int[100];
int ans = 0;
for (var e : dominoes) {
int x = e[0] < e[1] ? e[0] * 10 + e[1] : e[1] * 10 + e[0];
ans += cnt[x]++;
}
return ans;
}
}class Solution {
public:
int numEquivDominoPairs(vector<vector<int>>& dominoes) {
int cnt[100]{};
int ans = 0;
for (auto& e : dominoes) {
int x = e[0] < e[1] ? e[0] * 10 + e[1] : e[1] * 10 + e[0];
ans += cnt[x]++;
}
return ans;
}
};func numEquivDominoPairs(dominoes [][]int) (ans int) {
cnt := [100]int{}
for _, e := range dominoes {
x := e[0]*10 + e[1]
if e[0] > e[1] {
x = e[1]*10 + e[0]
}
ans += cnt[x]
cnt[x]++
}
return
}