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.

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

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.