Find Players With Zero or One Losses

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Arrays

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 <= 10⁵
matches[i].length == 2
1 <= winner_i, loser_i <= 10⁵
winner_i != loser_i
All matches[i] are unique.

Try coding it in LeetCode

Find Players With Zero or One Losses

Medium

Asked by:

15 views

Topics:

Arrays

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 <= 10⁵
matches[i].length == 2
1 <= winner_i, loser_i <= 10⁵
winner_i != loser_i
All matches[i] are unique.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible player IDs, and what data type are they (e.g., integer)?
Can the 'matches' input array be empty, or can a player appear multiple times as a winner or loser?
If a player has never won but has one or zero losses, should they be included in the output?
Should the output lists of players with zero losses and one loss be sorted in ascending order?
Are the player IDs guaranteed to be non-negative?

Brute Force Solution

Approach

We need to figure out which players lost zero games and which lost exactly one game based on a list of matches. The brute force method looks at each match individually and keeps track of how many times each player has lost.

Here's how the algorithm would work step-by-step:

Go through each match one at a time.
For each match, identify the player who lost (the one on the right side).
Keep a count of how many times each player appears as the loser across all the matches.
After going through all the matches, look at your counts and find the players who never lost (their count is zero).
Then, find the players who lost only once (their count is one).
List the players with zero losses and the players with one loss as your answer.

Code Implementation

def find_players_with_zero_or_one_losses(matches):
    losses_count = {}

    # Count losses for each player
    for match in matches:
        winner, loser = match[0], match[1]
        if loser not in losses_count:
            losses_count[loser] = 0
        losses_count[loser] += 1

        if winner not in losses_count:
            losses_count[winner] = 0

    no_losses = []
    one_loss = []

    # Segregate players based on losses
    for player, loss_count in losses_count.items():
        if loss_count == 0:
            no_losses.append(player)

        elif loss_count == 1:
            one_loss.append(player)

    no_losses.sort()
    one_loss.sort()

    # Need to return players with 0 and 1 losses in separate lists
    return [no_losses, one_loss]

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the list of matches once, where n is the number of matches. During each iteration, it updates the loss counts for a player. After processing all matches, it iterates through the loss counts to identify players with zero or one losses. Therefore the time complexity is driven by the single pass through the match data, resulting in O(n) time complexity.

Space Complexity

O(M) – The algorithm uses a data structure (likely a hash map or an array) to store the number of losses for each player. In the worst case, every player in every match could be a unique player, so the size of this data structure is proportional to the number of unique players which we'll represent as M. Therefore, the auxiliary space used is proportional to M. Thus, the space complexity is O(M).

Optimal Solution

Approach

We want to efficiently determine which players have lost zero or one matches. The optimal approach involves counting how many losses each player has incurred, then filtering the results to identify those with zero or one loss.

Here's how the algorithm would work step-by-step:

First, create a record to keep track of how many times each player has lost a match.
Go through the list of matches, and for each match, record a loss for the losing player.
Once you have counted the losses for everyone, create a list of players who haven't lost any matches (zero losses).
Create another list of players who have lost exactly one match.
Finally, present these two lists as the result, making sure to arrange them in the desired order.

Code Implementation

def find_players_with_zero_or_one_losses(matches):
    losses_count = {}

    for winner, loser in matches:
        losses_count.setdefault(winner, 0)
        losses_count[loser] = losses_count.get(loser, 0) + 1

    # Initialize lists to store players with zero and one losses.
    no_loss_players = []
    one_loss_players = []

    for player, loss_count in losses_count.items():
        if loss_count == 0:
            no_loss_players.append(player)
        elif loss_count == 1:
            one_loss_players.append(player)

    # Ensure that the lists are sorted.
    no_loss_players.sort()
    one_loss_players.sort()

    # Return the result as a list of lists.
    return [no_loss_players, one_loss_players]

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the 'matches' list (n matches) once to count losses for each player using a hash map (or similar data structure), resulting in O(n) time complexity. Subsequently, iterating through the hash map to gather players with zero and one losses also takes O(n) time, where n refers to the number of unique players, which is bounded by the number of matches. Finally, sorting the two result lists (at most n elements combined) contributes O(n log n), but this is still dominated by the initial linear time count. Therefore, the dominant term remains O(n) for iterating through the matches and O(n) to gather the lists so the overall runtime is O(n).

Space Complexity

O(N) – The algorithm uses a hash map (or dictionary) to store the number of losses for each player. In the worst case, every player in every match is unique, requiring space to store N players' loss counts, where N is the total number of distinct players across all matches. Additionally, two lists are created to store players with zero losses and one loss, respectively, which, in the worst case, could also hold up to N players. Therefore, the auxiliary space used is proportional to the number of unique players, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Empty matches array	Return two empty lists because there are no players or losses to process.
All players have zero losses	The 'zero losses' list will contain all players; the 'one loss' list will be empty.
All players have more than one loss	Both the 'zero losses' and 'one loss' lists will be empty.
A player appears as a loser but not as a winner	The player should be added to the set of all players and treated normally when counting losses.
Large number of matches causing potential memory issues	Use efficient data structures like hash maps or sets to minimize memory usage.
Integer overflow when counting losses for a specific player (highly unlikely, but consider)	Consider using a larger integer type if the problem description allows for a massive number of losses.
Matches array contains duplicate match entries	The number of losses for a player is still counted correctly because we iterate through all match entries.
Maximum number of players, and performance is crucial	Ensure that algorithms used have optimal time complexity (e.g., O(n log n) or O(n) where n is the number of matches).

Empty matches array

How to Handle:

Return two empty lists because there are no players or losses to process.

All players have zero losses

How to Handle:

The 'zero losses' list will contain all players; the 'one loss' list will be empty.

All players have more than one loss

How to Handle:

Both the 'zero losses' and 'one loss' lists will be empty.

A player appears as a loser but not as a winner

How to Handle:

The player should be added to the set of all players and treated normally when counting losses.

Large number of matches causing potential memory issues

How to Handle:

Use efficient data structures like hash maps or sets to minimize memory usage.

Integer overflow when counting losses for a specific player (highly unlikely, but consider)

How to Handle:

Consider using a larger integer type if the problem description allows for a massive number of losses.

Matches array contains duplicate match entries

How to Handle:

The number of losses for a player is still counted correctly because we iterate through all match entries.

Maximum number of players, and performance is crucial

How to Handle:

Ensure that algorithms used have optimal time complexity (e.g., O(n log n) or O(n) where n is the number of matches).