LeetCode in Kotlin

2225. Find Players With Zero or One Losses

Medium

You are given an integer array matches where matches[i] = [winneri, loseri] indicates that the player winneri defeated player loseri in a match.

Return a list answer of size 2 where:

• answer[0] is a list of all players that have not lost any matches.
• answer[1] is a list of all players that have lost exactly one match.

The values in the two lists should be returned in increasing order.

Note:

• You should only consider the players that have played at least one match.
• The testcases will be generated such that no two matches will have the same outcome.

Example 1:

Input: matches = [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]]

Output: [[1,2,10],[4,5,7,8]]

Explanation: Players 1, 2, and 10 have not lost any matches. Players 4, 5, 7, and 8 each have lost one match. Players 3, 6, and 9 each have lost two matches. Thus, answer[0] = [1,2,10] and answer[1] = [4,5,7,8].

Example 2:

Input: matches = [[2,3],[1,3],[5,4],[6,4]]

Output: [[1,2,5,6],[]]

Explanation: Players 1, 2, 5, and 6 have not lost any matches. Players 3 and 4 each have lost two matches. Thus, answer[0] = [1,2,5,6] and answer[1] = [].

Constraints:

• 1 <= matches.length <= 105
• matches[i].length == 2
• 1 <= winneri, loseri <= 105
• winneri != loseri
• All matches[i] are unique.

Solution

class Solution {
    fun findWinners(matches: Array<IntArray>): List<List<Int>> {
        val map: MutableMap<Int, Int> = HashMap()
        val list1: MutableList<Int> = ArrayList()
        val list2: MutableList<Int> = ArrayList()
        val res: MutableList<List<Int>> = ArrayList()
        for (match in matches) {
            val winner = match[0]
            val loser = match[1]
            map.putIfAbsent(winner, 0)
            map[loser] = map.getOrDefault(loser, 0) + 1
        }
        for ((key, value) in map) {
            if (value == 0) {
                list1.add(key)
            }
            if (value == 1) {
                list2.add(key)
            }
        }
        list1.sort()
        list2.sort()
        res.add(list1)
        res.add(list2)
        return res
    }
}

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