How do you solve Find Players With Zero or One Losses on LeetCode?

Find Players With Zero or One Losses (LeetCode #2225) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient counting of losses without needing nested loops.

What algorithm or data structure is used to solve Find Players With Zero or One Losses?

Find Players With Zero or One Losses uses the following concepts: Array, Hash Table, Sorting, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Players With Zero or One Losses?

Always clarify the requirements and constraints of the problem before jumping into coding. Think about edge cases, such as when there are no matches or when all players have lost. Explain your thought process and approach as you code, as this demonstrates your problem-solving skills to the interviewer.

What are common mistakes when solving Find Players With Zero or One Losses?

Failing to initialize the hash map correctly, leading to incorrect counts. Not considering players who have not lost any matches, which can lead to missing entries in the output.

Find Players With Zero or One Losses — LeetCode #2225 (Medium)

Tags: Array, Hash Table, Sorting, Counting

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

In the brute force approach, we will iterate through the matches and for each player, we will count how many matches they have lost. This is straightforward but inefficient, as it requires checking each player against all matches.

The time complexity is O(n²) because for each unique player, we iterate through all matches to count losses. This can be slow for large inputs.

Step 1: Create a list of unique players from the matches.
Step 2: For each player, count the number of losses by iterating through the matches.
Step 3: Classify players into two lists based on their loss count: zero losses and one loss.

Input: [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]] 1. Players set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 2. Count losses for player 1: 0 losses 3. Count losses for player 2: 0 losses 4. Count losses for player 3: 2 losses 5. Count losses for player 4: 3 losses 6. Count losses for player 5: 1 loss 7. Count losses for player 6: 2 losses 8. Count losses for player 7: 0 losses 9. Count losses for player 8: 0 losses 10. Count losses for player 9: 2 losses 11. Count losses for player 10: 0 losses Final output: [[1, 2, 10], [4, 5, 7, 8]]

Optimal Solution approach

Time complexity: O(n). Space complexity: O(n).

In the optimal solution, we use a single pass through the matches to count losses for each player using a hash map. This reduces the time complexity significantly, allowing us to efficiently categorize players.

The time complexity is O(n) because we only iterate through the matches once and then through the hash map, which is efficient for large inputs.

Step 1: Initialize a hash map to count losses for each player.
Step 2: Iterate through the matches and update the loss count for the loser in the hash map.
Step 3: Create two lists for players with zero losses and one loss by checking the hash map.

Input: [[1,3],[2,3],[3,6],[5,6],[5,7],[4,5],[4,8],[4,9],[10,4],[10,9]] 1. Initialize lossCount: {} 2. After processing [1,3]: lossCount: {3: 1, 1: 0} 3. After processing [2,3]: lossCount: {3: 2, 1: 0, 2: 0} 4. After processing [3,6]: lossCount: {3: 2, 1: 0, 2: 0, 6: 1} 5. After processing [5,6]: lossCount: {3: 2, 1: 0, 2: 0, 6: 1, 5: 0} 6. After processing [5,7]: lossCount: {3: 2, 1: 0, 2: 0, 6: 1, 5: 0, 7: 0} Final output: [[1, 2, 10], [4, 5, 7, 8]]

Key Insights

Using a hash map allows for efficient counting of losses without needing nested loops.
Sorting the final lists ensures the output is in increasing order, which is a common requirement.

Common Mistakes

Failing to initialize the hash map correctly, leading to incorrect counts.
Not considering players who have not lost any matches, which can lead to missing entries in the output.

Interview Tips

Always clarify the requirements and constraints of the problem before jumping into coding.
Think about edge cases, such as when there are no matches or when all players have lost.
Explain your thought process and approach as you code, as this demonstrates your problem-solving skills to the interviewer.

#2225

Find Players With Zero or One Losses

Medium

Array↗Hash Table↗Sorting↗Counting↗Hash Map↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

⚙️

Algorithm

3 steps

1Step 1: Initialize a hash map to count losses for each player.
2Step 2: Iterate through the matches and update the loss count for the loser in the hash map.
3Step 3: Create two lists for players with zero losses and one loss by checking the hash map.

solution.py8 lines

1def findWinners(matches):
2    loss_count = {}
3    for winner, loser in matches:
4        loss_count[loser] = loss_count.get(loser, 0) + 1
5        loss_count.setdefault(winner, 0)
6    zero_losses = [player for player, losses in loss_count.items() if losses == 0]
7    one_loss = [player for player, losses in loss_count.items() if losses == 1]
8    return [sorted(zero_losses), sorted(one_loss)]

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the matches once and then through the hash map, which is efficient for large inputs.

1Using a hash map allows for efficient counting of losses without needing nested loops.
2Sorting the final lists ensures the output is in increasing order, which is a common requirement.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.