2225. Find Players With Zero or One Losses

MediumArrayHash TableCountingSorting
Leetcode Link

Problem Description

The problem provides you with an integer array matches, where each element represents the outcome of a match as [winner_i, loser_i]. The task is to find all players who have not lost any matches and all players who have lost exactly one match, ensuring that all listed players have participated in at least one match.

Here's how you need to organize the output:

  • The first list (answer[0]) should contain all players who have never lost a match.
  • The second list (answer[1]) should include all players who have lost exactly one match.

It's important to list the players in ascending order. If a player does not appear in the matches array, they are not considered since no match records for them exist. Additionally, players will not lose more than one match because match outcomes are unique.

Intuition

The intuition behind the solution is to track the number of losses for each player. This is achievable by counting the occurrences of each player in the loser position of every match.

To do this systematically, we can use the following steps:

  1. Utilize a Counter to keep a tally of the losses for each player.
  2. Iterate over the matches list, incrementing the count for losers. If a player wins a match, ensure they're in the Counter with a loss count of 0 (since winning means they have not lost that match).
  3. After processing all matches, go through the Counter and categorize players based on their loss count.
    • Players with 0 losses go into answer[0].
    • Players with exactly 1 loss go into answer[1].
  4. Sort both lists to satisfy the ascending order requirement.
  5. Return the sorted lists as the final result.

Using the Counter, we abstract the complexity of tracking individual losses and make it simple to determine which list a player belongs to based on their number of losses after processing all match outcomes.

Learn more about Sorting patterns.

Solution Approach

The solution's implementation follows an efficient approach to categorize players based on their match outcomes:

  1. A Counter object is utilized to maintain the tally of losses for each player. This data structure is optimal for this purpose because it allows us to keep track of how many times each player appears as the loser and does not count their wins.

  2. Iterate through the matches list, and for each match [winner, loser]:

    • Check if the winner is already in the Counter. If not, initialize their loss count to 0 because winning a match implies they haven't lost it.
    • Increment the loser's count by one to signify their loss in this match.
  3. After tallying the losses for each player, create a two-dimensional list ans, where ans[0] will eventually contain players with 0 losses, and ans[1] will contain players with exactly one loss.

  4. Iterate through the items in the Counter (for u, v in cnt.items():). For each player (u) and their loss count (v):

    • If v is less than 2 (meaning the player has either won all their matches or lost just one), add the player to ans[v]. This works because v can only be 0 or 1, as we are only interested in players with no losses or exactly one loss.
  5. The lists need to be sorted in increasing order, as per the problem specifications. Thus, both ans[0] and ans[1] are sorted using the sort() method.

  6. Finally, return the list ans as the result, where ans[0] contains all players that have not lost any matches and ans[1] contains all players that have lost exactly one match.

This algorithm efficiently uses a hash map (provided by Counter in Python) to count occurrences, which is ideal for frequency counting tasks. The overall time complexity of the algorithm is determined by the number of matches and the sorting step, and the space complexity is largely influenced by the Counter used to store losses per player.

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Example Walkthrough

Let's go through a small example to illustrate the solution approach. Assume we have the following matches array:

matches = [[1, 3], [2, 1], [4, 2], [5, 2]]

In this array, the subarrays represent match outcomes with winners and losers respectively.

  1. Initialize a Counter to count losses:

    • After initializing: Counter({})
  2. Start iterating through matches:

    • Match [1, 3]: Counter({'3': 1})
    • Match [2, 1]: Counter({'3': 1, '1': 1})
    • Before proceeding with the loser of the next match, ensure that winner 4 has a loss count of 0: Counter({'3': 1, '1': 1, '4': 0})
    • Match [4, 2]: Counter({'3': 1, '1': 1, '2': 1, '4': 0})
    • Again, ensure that winner 5 has a loss count of 0 before the next loser is processed: Counter({'3': 1, '1': 1, '2': 1, '4': 0, '5': 0})
    • Match [5, 2]: Counter({'3': 1, '1': 1, '2': 2, '4': 0, '5': 0})
  3. Now we have all loss counts, create list ans with two sublists for 0 losses and 1 loss:

    • ans = [[], []]
  4. Iterate through the counts and populate ans accordingly:

    • Player 3 has 1 loss: ans = [[], [3]]
    • Player 1 has 1 loss: ans = [[], [3, 1]]
    • Player 4 has 0 losses: ans = [[4], [3, 1]]
    • Player 5 has 0 losses: ans = [[4, 5], [3, 1]]
    • Player 2 is not added to either sublist since they've lost more than one match.
  5. Sort both ans[0] and ans[1] as required:

    • ans[0].sort(): ans = [[4, 5], [3, 1]]
    • ans[1].sort(): ans = [[4, 5], [1, 3]]
  6. The final sorted ans list is returned from the function:

ans = [[4, 5], [1, 3]]

This holds our result, where ans[0] contains players 4 and 5 who have never lost any matches, and ans[1] contains players 1 and 3 who have lost exactly one match, listed in ascending order.

Solution Implementation

1from collections import Counter
2
3class Solution:
4    def findWinners(self, matches):
5        count_losses = Counter()
6
7        # Count the number of losses for each player
8        for winner, loser in matches:
9            # Ensure winners with no losses are accounted for
10            if winner not in count_losses:
11                count_losses[winner] = 0
12
13            # Increment the loss count for losers
14            count_losses[loser] += 1
15
16        winners_with_zero_losses = []
17        winners_with_one_loss = []
18
19        # Iterate over the players and their loss counts
20        for player, loss_count in count_losses.items():
21            # If a player has 0 losses, add them to the winners_with_zero_losses list
22            if loss_count == 0:
23                winners_with_zero_losses.append(player)
24            # If a player has exactly 1 loss, add them to the winners_with_one_loss list
25            elif loss_count == 1:
26                winners_with_one_loss.append(player)
27
28        # Sort the lists to meet the output criteria
29        winners_with_zero_losses.sort()
30        winners_with_one_loss.sort()
31
32        # Combine the two lists into a single list of lists for the result
33        result = [winners_with_zero_losses, winners_with_one_loss]
34        return result
35
1import java.util.ArrayList;
2import java.util.HashMap;
3import java.util.List;
4import java.util.Map;
5import java.util.Collections;
6
7class Solution {
8    public List<List<Integer>> findWinners(int[][] matches) {
9        // A map to keep track of loss counts for each player
10        Map<Integer, Integer> lossCountMap = new HashMap<>();
11        // Process all match results
12        for (int[] match : matches) {
13            int winner = match[0];
14            int loser = match[1];
15            // Initialize the winner's loss count if not already present in map
16            lossCountMap.putIfAbsent(winner, 0);
17            // Increment the loss count for the loser
18            lossCountMap.put(loser, lossCountMap.getOrDefault(loser, 0) + 1);
19        }
20
21        // Create the result list containing two lists: one for all the players who have never lost, and one for the players who have lost exactly once
22        List<List<Integer>> winnersList = new ArrayList<>();
23        winnersList.add(new ArrayList<>());
24        winnersList.add(new ArrayList<>());
25
26
27        // Iterate through each entry in the loss count map
28        for (Map.Entry<Integer, Integer> entry : lossCountMap.entrySet()) {
29            int player = entry.getKey();
30            int losses = entry.getValue();
31            // If the player has lost fewer than 2 matches, include them in the appropriate sublist (0 losses or 1 loss)
32            if (losses < 2) {
33                winnersList.get(losses).add(player);
34            }
35        }
36
37        // Sort the sublists of players with 0 losses and exactly 1 loss
38        Collections.sort(winnersList.get(0));
39        Collections.sort(winnersList.get(1));
40
41        // Return the result list
42        return winnersList;
43    }
44}
45
1#include <vector>
2#include <unordered_map>
3#include <algorithm>
4
5class Solution {
6public:
7    // Function to find the players who have never lost a match (winners)
8    // and those who have lost exactly one match (one-match-losers)
9    vector<vector<int>> findWinners(vector<vector<int>>& matches) {
10        unordered_map<int, int> lossCount; // Map storing the loss count for each player
11
12        // Process each match result and update the loss count for the players
13        for (auto& match : matches) {
14            int winner = match[0];
15            int loser = match[1];
16
17            // Make sure every player is included in the map
18            if (!lossCount.count(winner)) {
19                lossCount[winner] = 0;
20            }
21
22            // Increment the loss count for the loser of the match
23            ++lossCount[loser];
24        }
25
26        vector<vector<int>> answer(2); // To hold the final result
27
28        // Iterate through the map to classify players based on their loss counts
29        for (auto& playerLossPair : lossCount) {
30            int player = playerLossPair.first;
31            int losses = playerLossPair.second;
32
33            // If the player has less than 2 losses, add them to the respective list
34            if (losses < 2) {
35                answer[losses].push_back(player);
36            }
37        }
38
39        // Sort the list of winners and one-match-losers
40        sort(answer[0].begin(), answer[0].end());
41        sort(answer[1].begin(), answer[1].end());
42
43        return answer; // Return the sorted lists
44    }
45};
46
1// Import statements are not required in TypeScript for data structures like arrays and maps.
2
3// Type Alias for readability, representing a match result with winner and loser.
4type MatchResult = [number, number];
5
6// Function to update the loss count map with match results
7function processMatch(lossCount: Map<number, number>, winner: number, loser: number): void {
8  // Ensure every player is present in the map; if not add them with zero losses
9  if (!lossCount.has(winner)) {
10    lossCount.set(winner, 0);
11  }
12
13  // Increment the loss count for the loser of the match
14  const currentLossCount = lossCount.get(loser) || 0;
15  lossCount.set(loser, currentLossCount + 1);
16}
17
18// Function to sort and categorize players based on their loss counts into winners and one-match-losers
19function findWinners(matches: MatchResult[]): number[][] {
20  const lossCount: Map<number, number> = new Map(); // Map to store the loss count for each player
21
22  // Process each match result and update the loss count for the players
23  for (const match of matches) {
24    const [winner, loser] = match;
25    processMatch(lossCount, winner, loser);
26  }
27
28  // Arrays to hold the list of players who never lost and those who lost exactly one match
29  const winners: number[] = [];
30  const oneMatchLosers: number[] = [];
31
32  // Iterate through the map to classify players based on their loss counts
33  for (const [player, losses] of lossCount) {
34    // If the player has no losses, they are a winner. If only one loss, they are a one-match-loser.
35    if (losses === 0) {
36      winners.push(player);
37    } else if (losses === 1) {
38      oneMatchLosers.push(player);
39    }
40  }
41
42  // Sort the list of winners and one-match-losers
43  winners.sort((a, b) => a - b);
44  oneMatchLosers.sort((a, b) => a - b);
45
46  // Return the sorted lists as a 2D array
47  return [winners, oneMatchLosers];
48}
49
50// Usage of the findWinners function can be demonstrated with an example:
51const matchResults: MatchResult[] = [
52  [1, 3],
53  [2, 3],
54  [3, 6],
55  [5, 6],
56  [5, 7]
57];
58const results = findWinners(matchResults);
59console.log(`Winners: ${results[0]}, One-Match Losers: ${results[1]}`);
60

Time and Space Complexity

The given Python code seeks to find all players who never lost a game (who are undefeated), and all players who lost exactly one game. Let's break down its time and space complexity.

Time Complexity

  1. Creating the counter: O(N) - Creating the counter object requires iterating over the list of matches where N is the number of matches.

  2. Initializing player counts: O(N) - We iterate through all matches, and for each match we perform a check and counter increment which operates in constant time, resulting in O(N).

  3. Traversing the counter for players with losses less than 2: O(P) - We go through the counter which contains P unique players.

  4. Sorting winners and players with one loss: O(PlogP) - Sorting is performed on the players' list for those who have not lost and those who have lost one match. In the worst case, all players could either be winners or have one loss, hence the sorting can be O(PlogP) for P unique players.

The overall time complexity is the sum of these operations, dominated by the sorting steps: O(N) + O(P) + O(PlogP). Since the sorting term is usually the most significant for large lists, we can simplify this expression to: O(PlogP).

Space Complexity

  1. Counter object: O(P) - The counter object holds at most P unique players, which is the space required.

  2. Answer List: O(P) - The answer list is a list of two lists, which in the worst case would hold all P players, resulting in O(P) space.

The overall space complexity combines both aspects, remaining O(P) since it is not multiplied by any factor.

Therefore, the final complexity of the provided code is:

  • Time Complexity: O(PlogP)
  • Space Complexity: O(P)

Learn more about how to find time and space complexity quickly using problem constraints.


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