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.

Approach 1: Hash Map (1062ms)

Check the number of times to lose for each player. If it is $0$ , then this player belongs to $ans[0]$ . If it is $1$ , then it belongs to $ans[1]$ .

Let $n$ be the number of matches then the

Time complexity is $O(n \log n)$ we need to look at all the matches which would be $O(n)$ but the sort at the end is $O(n \log n)$ (kudos @learner_9 for spotting and pointing out the mistake) and the
Space complexity is $O(n)$ .

class Solution {
public:
    vector<vector<int>> findWinners(vector<vector<int>>& matches) {
        vector<vector<int>> ans(2);
        unordered_map<int, int> lost;
        // calculate lost matches
        for (auto x : matches) {
            if (!lost.count(x[0])) lost[x[0]] = 0;
            if (!lost.count(x[1])) lost[x[1]] = 0;
            lost[x[1]]++;
        }
        // categorise players
        for (auto x : lost) {
            int who = x.first, lostMatches = x.second;
            if (lostMatches == 0) ans[0].push_back(who);
            else if (lostMatches == 1) ans[1].push_back(who);
        }
        // the values in the two lists should be returned in increasing order
        sort(ans[0].begin(), ans[0].end());
        sort(ans[1].begin(), ans[1].end());
        return ans;
    }
};

Approach 2: Set (1249 ms)

Let $n$ be the number of matches then the

Time complexity is $O(n \log n)$ because of the insertion into the set / map and the
Space complexity is $O(n)$ .

static vector<vector<int>> findWinners(const vector<vector<int>>& matches) noexcept {
    vector<vector<int>> ans(2);

    set<int> win;
    map<int, int> loss;

    for (const vector<int>& match : matches) {
        win.insert(match[0]);
        ++loss[match[1]];
    }

    for (int p : win)
        if (loss.find(p) == end(loss)) ans[0].push_back(p);

    for (auto [p, l] : loss)
        if (l == 1) ans[1].push_back(p);

    return ans;
}

Variant with a single map (1069ms)

Picking up on a idea from @stanislav-iablokov we can turn this into a solution with only one map. This is as fast as the solution with the hash set / map.

static vector<vector<int>> findWinners(const vector<vector<int>>& matches) noexcept {
    vector<vector<int>> ans(2);

    map<int, int> loss;

    for (const vector<int>& match : matches) {
        loss[match[0]] += 0;  // make sure the winners are in the map too., just  loss[match[0]]; would work as well
        ++loss[match[1]];
    }

    for (auto [p, l] : loss) {
        if (l == 0) {
            ans[0].push_back(p);
        } else if (l == 1) {
            ans[1].push_back(p);
        }
    }

    return ans;
}

Approach 3: Arrays (749ms)

The range of the play number is limited enough that we can just use arrays instead of a hash map / set.

static vector<vector<int>> findWinners(const vector<vector<int>>& matches) noexcept {
    array<bool, 100001> played = {};
    array<int, 100001> losses = {};

    for (const vector<int>& match : matches) {
        played[match[0]] = true;
        played[match[1]] = true;
        ++losses[match[1]];
    }

    vector<vector<int>> ans(2);

    for (int i = 0; i < size(played); ++i) {
        if (played[i]) {
            if (losses[i] == 0) {
                ans[0].push_back(i);
            } else if (losses[i] == 1) {
                ans[1].push_back(i);
            }
        }
    }

    return ans;
}

Let $n$ be the number of matches then the

Time complexity is $O(n + c)$ we need to look at all the matches, but there is a significat constant factor $c$ as we need look over the entier arrays and the
Space complexity is $O(c)$ , which means it's constant, but it's a large constant.

2225 - Find Players With Zero or One Losses (Medium)

Problem Link

Problem Statement

Approach 1: Hash Map (1062ms)

Approach 2: Set (1249 ms)

Variant with a single map (1069ms)

Approach 3: Arrays (749ms)

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