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Formatted question description: https://leetcode.ca/all/2225.html
2225. Find Players With Zero or One Losses (Medium)
You are given an integer array matches
where matches[i] = [winner_{i}, loser_{i}]
indicates that the player winner_{i}
defeated player loser_{i}
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 <= winner_{i}, loser_{i} <= 10^{5}
winner_{i} != loser_{i}
 All
matches[i]
are unique.
Companies:
Indeed
Related Topics:
Graph
Similar Questions:
Solution 1. Hash Set and Map
Traverse the array, store the unique winners in a set win
and store a mapping from person to lose count in map lose
.
Traverse lose
map, for each pair person, cnt
, add person
to vector oneLose
if cnt == 1
(i.e. this person loses exactly once), and erase this person
from win
.
Lastly, return the elements in win
and oneLose
.

// OJ: https://leetcode.com/problems/findplayerswithzerooronelosses/ // Time: O(NlogN) // Space: O(N) class Solution { public: vector<vector<int>> findWinners(vector<vector<int>>& A) { set<int> win; map<int, int> lose; // person, count for (auto &m : A) { win.insert(m[0]); lose[m[1]]++; } vector<int> oneLose; for (auto &[p, cnt] : lose) { if (cnt == 1) oneLose.push_back(p); win.erase(p); } return {vector<int>(begin(win), end(win)), oneLose}; } };

class Solution: def findWinners(self, matches: List[List[int]]) > List[List[int]]: cnt = Counter() for a, b in matches: if a not in cnt: cnt[a] = 0 cnt[b] += 1 ans = [[], []] for u, v in cnt.items(): if v < 2: ans[v].append(u) ans[0].sort() ans[1].sort() return ans ############ # 2225. Find Players With Zero or One Losses # https://leetcode.com/problems/findplayerswithzerooronelosses/ class Solution: def findWinners(self, matches: List[List[int]]) > List[List[int]]: win = defaultdict(int) lose = defaultdict(int) for x, y in matches: win[x] += 1 lose[y] += 1 res = [[] for _ in range(2)] for k, v in win.items(): if lose[k] == 0: res[0].append(k) for k, v in lose.items(): if v == 1: res[1].append(k) res[0].sort() res[1].sort() return res

class Solution { public List<List<Integer>> findWinners(int[][] matches) { Map<Integer, Integer> cnt = new HashMap<>(); for (int[] m : matches) { int a = m[0], b = m[1]; cnt.putIfAbsent(a, 0); cnt.put(b, cnt.getOrDefault(b, 0) + 1); } List<List<Integer>> ans = new ArrayList<>(); ans.add(new ArrayList<>()); ans.add(new ArrayList<>()); for (Map.Entry<Integer, Integer> entry : cnt.entrySet()) { int u = entry.getKey(); int v = entry.getValue(); if (v < 2) { ans.get(v).add(u); } } Collections.sort(ans.get(0)); Collections.sort(ans.get(1)); return ans; } }

func findWinners(matches [][]int) [][]int { cnt := map[int]int{} for _, m := range matches { a, b := m[0], m[1] if _, ok := cnt[a]; !ok { cnt[a] = 0 } cnt[b]++ } ans := make([][]int, 2) for u, v := range cnt { if v < 2 { ans[v] = append(ans[v], u) } } sort.Ints(ans[0]) sort.Ints(ans[1]) return ans }
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https://leetcode.com/problems/findplayerswithzerooronelosses/discuss/1908802/