LeetCode in Kotlin

1128. Number of Equivalent Domino Pairs

Easy

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 * 104
• dominoes[i].length == 2
• 1 <= dominoes[i][j] <= 9

Solution

class Solution {
    fun numEquivDominoPairs(dominoes: Array<IntArray>): Int {
        val map: MutableMap<Int, Int> = HashMap()
        var count = 0
        for (dominoe in dominoes) {
            val smaller = Math.min(dominoe[0], dominoe[1])
            val bigger = Math.max(dominoe[0], dominoe[1])
            val key = smaller * 10 + bigger
            count += map.getOrDefault(key, 0)
            map[key] = map.getOrDefault(key, 0) + 1
        }
        return count
    }
}

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