Formatted question description: https://leetcode.ca/all/1128.html

1128. Number of Equivalent Domino Pairs (Easy)

Given a list of dominoesdominoes[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

 

Constraints:

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

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/number-of-equivalent-domino-pairs/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int numEquivDominoPairs(vector<vector<int>>& A) {
        map<vector<int>, int> m;
        int ans = 0;
        for (auto &v : A) {
            sort(begin(v), end(v));
            ans += m[v];
            m[v]++;
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/number-of-equivalent-domino-pairs/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int numEquivDominoPairs(vector<vector<int>>& A) {
        for (auto &v : A) sort(begin(v), end(v));
        sort(begin(A), end(A));
        int ans = 0;
        for (int i = 0, N = A.size(); i < N;) {
            int start = i;
            while (i < N && A[i] == A[start]) ++i;
            int len = i - start;
            ans += (len - 1) * len / 2;
        }
        return ans;
    }
};

Solution 3.

// OJ: https://leetcode.com/problems/number-of-equivalent-domino-pairs/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numEquivDominoPairs(vector<vector<int>>& A) {
        int cnt[10][10] = {}, ans = 0;
        for (auto &v : A) {
            if (v[0] > v[1]) swap(v[0], v[1]);
            ans += cnt[v[0]][v[1]]++;
        }
        return ans;
    }
};

Java

  • class Solution {
        public int numEquivDominoPairs(int[][] dominoes) {
            int length = dominoes.length;
            for (int i = 0; i < length; i++) {
                int[] domino = dominoes[i];
                if (domino[0] > domino[1]) {
                    int temp = domino[0];
                    dominoes[i][0] = dominoes[i][1];
                    dominoes[i][1] = temp;
                }
            }
            Arrays.sort(dominoes, new Comparator<int[]>() {
                public int compare(int[] array1, int[] array2) {
                    if (array1[0] != array2[0])
                        return array1[0] - array2[0];
                    else
                        return array1[1] - array2[1];
                }
            });
            int equivalentCount = 0;
            int consecutive = 1;
            for (int i = 1; i < length; i++) {
                int[] prevDomino = dominoes[i - 1];
                int[] curDomino = dominoes[i];
                if (prevDomino[0] == curDomino[0] && prevDomino[1] == curDomino[1])
                    consecutive++;
                else {
                    equivalentCount += consecutive * (consecutive - 1) / 2;
                    consecutive = 1;
                }
            }
            if (consecutive > 1)
                equivalentCount += consecutive * (consecutive - 1) / 2;
            return equivalentCount;
        }
    }
    
  • // OJ: https://leetcode.com/problems/number-of-equivalent-domino-pairs/
    // Time: O(NlogN)
    // Space: O(N)
    class Solution {
    public:
        int numEquivDominoPairs(vector<vector<int>>& A) {
            map<vector<int>, int> m;
            int ans = 0;
            for (auto &v : A) {
                sort(begin(v), end(v));
                ans += m[v];
                m[v]++;
            }
            return ans;
        }
    };
    
  • class Solution(object):
        def numEquivDominoPairs(self, dominoes):
            count = dict()
            for domi in dominoes:
                hash_ = domi[0] * 10 + domi[1] if domi[0] < domi[1] else domi[1] * 10 + domi[0]
                if hash_ in count:
                    count[hash_] += 1
                else:
                    count[hash_] = 1
            res = 0
            for v in count.values():
                res += v * (v - 1) / 2
            return res
    

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