Number of Equivalent Domino Pairs - Array - Easy - LeetCode

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

Constraints:

1 <= dominoes.length <= 40000

1 <= dominoes[i][j] <= 9

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.Array
{
 public class NumEquivDominoPairsSoln
 {
 public int NumEquivDominoPairs(int[][] dominoes)
 {
 var map = new Dictionary<string, int>();
 int count = 0;
 string key = string.Empty;
 for (int i = 0; i < dominoes.Length; i++)
 {
 key = Math.Min(dominoes[i][0], dominoes[i][1]).ToString() +
 "-" +Math.Max(dominoes[i][0], dominoes[i][1]).ToString();

 if (map.ContainsKey(key))
 {
 count += map[key];
 map[key]++;
 }
 else
 {
 map.Add(key, 1);
 } 
 }

 return count;
 } 
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

Unit Tests:

using LeetCode.AskGif.Easy.Array;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;

namespace CodingUnitTest.Easy.Array
{
 [TestClass]
 public class NumEquivDominoPairsSolnTests
 {
 [TestMethod]
 public void NumEquivDominoPairsSoln_First()
 {
 var dominoes = new int[,]{
 { 1, 2 },
 { 2, 1 },
 { 3, 4 },
 { 5, 6 }
 }; 
 var expected = 1;

 var res = new NumEquivDominoPairsSoln().NumEquivDominoPairs(ArrayMapper(dominoes));
 Assert.AreEqual(expected, res);
 }

 [TestMethod]
 public void NumEquivDominoPairsSoln_Second()
 {
 var dominoes = new int[,]{
 { 1, 1 },
 { 2, 2 },
 { 1, 1 },
 { 1, 2 },
 { 1, 2 },
 { 1, 1 }
 };
 var expected = 4;

 var res = new NumEquivDominoPairsSoln().NumEquivDominoPairs(ArrayMapper(dominoes));
 Assert.AreEqual(expected, res);
 }

 private int[][] ArrayMapper(int[,] matrix)
 {
 var arr = new int[matrix.GetLength(0)][];
 for (int i = 0; i < matrix.GetLength(0); i++)
 {
 arr[i] = new int[matrix.GetLength(1)];
 for (int j = 0; j < matrix.GetLength(1); j++)
 {
 arr[i][j] = matrix[i, j];
 }
 }

 return arr;
 }
 }
}