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2225. Find Players With Zero or One Losses

Description

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.

Solutions

Solution 1: Hash Table + Sorting

Thinking

We must list players who never lost and those who lost exactly once, each in increasing order. There are up to \(10^5\) matches, so scanning every possible id is wasteful. Only players who appear matter, and the only statistic is the loss count.

A hash map \(\textit{cnt}\) stores losses: a winner is inserted with \(0\) if new, a loser is incremented. After sorting by id, players with \(0\) or \(1\) loss go into the two answer lists.

We use a hash table cnt to record the number of matches each player has lost.

Then we traverse the hash table, put the players who lost 0 matches into ans[0], and put the players who lost 1 match into ans[1].

Finally, we sort ans[0] and ans[1] in ascending order and return the result.

The time complexity is \(O(n \times \log n)\), and the space complexity is \(O(n)\). Where \(n\) is the number of matches.

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class Solution:
    def findWinners(self, matches: List[List[int]]) -> List[List[int]]:
        cnt = Counter()
        for winner, loser in matches:
            if winner not in cnt:
                cnt[winner] = 0
            cnt[loser] += 1
        ans = [[], []]
        for x, v in sorted(cnt.items()):
            if v < 2:
                ans[v].append(x)
        return ans

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class Solution {
    public List<List<Integer>> findWinners(int[][] matches) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (var e : matches) {
            cnt.putIfAbsent(e[0], 0);
            cnt.merge(e[1], 1, Integer::sum);
        }
        List<List<Integer>> ans = List.of(new ArrayList<>(), new ArrayList<>());
        for (var e : cnt.entrySet()) {
            if (e.getValue() < 2) {
                ans.get(e.getValue()).add(e.getKey());
            }
        }
        Collections.sort(ans.get(0));
        Collections.sort(ans.get(1));
        return ans;
    }
}

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class Solution {
public:
    vector<vector<int>> findWinners(vector<vector<int>>& matches) {
        map<int, int> cnt;
        for (auto& e : matches) {
            if (!cnt.contains(e[0])) {
                cnt[e[0]] = 0;
            }
            ++cnt[e[1]];
        }
        vector<vector<int>> ans(2);
        for (auto& [x, v] : cnt) {
            if (v < 2) {
                ans[v].push_back(x);
            }
        }
        return ans;
    }
};

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func findWinners(matches [][]int) [][]int {
    cnt := map[int]int{}
    for _, e := range matches {
        if _, ok := cnt[e[0]]; !ok {
            cnt[e[0]] = 0
        }
        cnt[e[1]]++
    }
    ans := make([][]int, 2)
    for x, v := range cnt {
        if v < 2 {
            ans[v] = append(ans[v], x)
        }
    }
    sort.Ints(ans[0])
    sort.Ints(ans[1])
    return ans
}

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function findWinners(matches: number[][]): number[][] {
    const cnt: Map<number, number> = new Map();
    for (const [winner, loser] of matches) {
        if (!cnt.has(winner)) {
            cnt.set(winner, 0);
        }
        cnt.set(loser, (cnt.get(loser) || 0) + 1);
    }
    const ans: number[][] = [[], []];
    for (const [x, v] of cnt) {
        if (v < 2) {
            ans[v].push(x);
        }
    }
    ans[0].sort((a, b) => a - b);
    ans[1].sort((a, b) => a - b);
    return ans;
}

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/**
 * @param {number[][]} matches
 * @return {number[][]}
 */
var findWinners = function (matches) {
    const cnt = new Map();
    for (const [winner, loser] of matches) {
        if (!cnt.has(winner)) {
            cnt.set(winner, 0);
        }
        cnt.set(loser, (cnt.get(loser) || 0) + 1);
    }
    const ans = [[], []];
    for (const [x, v] of cnt) {
        if (v < 2) {
            ans[v].push(x);
        }
    }
    ans[0].sort((a, b) => a - b);
    ans[1].sort((a, b) => a - b);
    return ans;
};

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