1128: Number-of-Equivalent-Domino-Pairs
Easy
You can just store the count of all the pairs in a
9x9 grid and calculate the number of pairs using the
n * (n-1) / 2 formula, where n denotes
the number of equivalent dominoes that exist.
Code:
class Solution {
public:
int numEquivDominoPairs(vector<vector<int>>& dominoes) {
vector<vector<int>> grid (9, vector<int>(9, 0));
for (auto& domino : dominoes) {
++grid[domino[0]-1][domino[1]-1];
}
int ans = 0;
for (int i = 0; i < 9; ++i) {
for (int j = i; j < 9; ++j) {
int total_pairs = grid[i][j];
if (i != j) {
total_pairs += grid[j][i];
}
ans += (total_pairs-1) * (total_pairs) / 2;
}
}
return ans;
}
};