Find Players With Zero or One Losses

Problem

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 <= 10^5
matches[i].length == 2
1 <= winneri, loseri <= 10^5
winneri != loseri
All matches[i] are unique.

Solution (Java)

class Solution {
    public List<List<Integer>> findWinners(int[][] matches) {
        Map<Integer, Integer> map = new HashMap<>();
        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();
        List<List<Integer>> res = new ArrayList<>();
        for (int[] match : matches) {
            int winner = match[0];
            int loser = match[1];
            map.putIfAbsent(winner, 0);
            map.put(loser, map.getOrDefault(loser, 0) + 1);
        }
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (entry.getValue() == 0) {
                list1.add(entry.getKey());
            }
            if (entry.getValue() == 1) {
                list2.add(entry.getKey());
            }
        }
        Collections.sort(list1);
        Collections.sort(list2);
        res.add(list1);
        res.add(list2);
        return res;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).