What is the time complexity of Find Players With Zero or One Losses?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Find Players With Zero or One Losses?

This problem uses Array, Hash Table, Sorting, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find Players With Zero or One Losses?

Insight 1: Use a hash map (or dictionary) to efficiently count the number of losses for each player. Insight 2: Iterate through the hash map to identify players with zero and one losses.

Which companies ask LeetCode 2225. Find Players With Zero or One Losses in interviews?

This problem is frequently asked at Indeed, Palantir Technologies.

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Find Players With Zero or One Losses - LeetCode 2225 Solution

Find Players With Zero or One Losses - Complete Solution Guide

Find Players With Zero or One Losses is LeetCode problem 2225, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 gener

Detailed Explanation

The problem asks us to analyze a list of matches between players and identify two groups of players: those who have never lost a match and those who have lost exactly one match. The input is a list of matches, where each match is represented as a pair [winner, loser]. We need to return a list containing two sorted lists: the first list containing players with zero losses, and the second containing players with one loss. We only consider players who participated in at least one match.

Solution Approach

The solution uses a hash map to store the number of losses for each player. We iterate through the input `matches` list, updating the loss count for each loser. If a winner isn't already in the hashmap, we add them with a loss count of 0. After processing all matches, we iterate through the hash map and categorize players based on their loss counts. Finally, we sort the lists of players with zero and one loss and return them in the specified format.

Step-by-Step Algorithm

Step 1: Initialize a hash map `losses_count` to store the number of losses for each player.
Step 2: Iterate through the `matches` list.
Step 3: For each match `[winner, loser]`, update the `losses_count` hash map. Increment the count for the `loser`. If the `winner` is not in the map, insert them with a count of 0.
Step 4: Initialize two empty lists, `zero_losses` and `one_loss`, to store players with zero and one loss respectively.
Step 5: Iterate through the `losses_count` hash map.
Step 6: For each player and their loss count, if the count is 0, add the player to the `zero_losses` list. If the count is 1, add the player to the `one_loss` list.
Step 7: Sort the `zero_losses` and `one_loss` lists in ascending order.
Step 8: Return a list containing the `zero_losses` and `one_loss` lists.

Key Insights

Insight 1: Use a hash map (or dictionary) to efficiently count the number of losses for each player.
Insight 2: Iterate through the hash map to identify players with zero and one losses.
Insight 3: Sort the resulting lists of players to meet the output requirements.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sorting, Counting.

Companies

Asked at: Indeed, Palantir Technologies.

Find Players With Zero or One Losses | LeetCode 2225

Medium

Array Hash Table Sorting Counting

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<List<Integer>> findWinners(int[][] matches) {
        Map<Integer, Integer> lossesCount = new HashMap<>();

        for (int[] match : matches) {
            int winner = match[0];
            int loser = match[1];

            if (!lossesCount.containsKey(winner)) {
                lossesCount.put(winner, 0);
            }

            lossesCount.put(loser, lossesCount.getOrDefault(loser, 0) + 1);
        }

        List<Integer> zeroLosses = new ArrayList<>();
        List<Integer> oneLoss = new ArrayList<>();

        for (Map.Entry<Integer, Integer> entry : lossesCount.entrySet()) {
            int player = entry.getKey();
            int count = entry.getValue();

            if (count == 0) {
                zeroLosses.add(player);
            } else if (count == 1) {
                oneLoss.add(player);
            }
        }

        Collections.sort(zeroLosses);
        Collections.sort(oneLoss);

        List<List<Integer>> result = new ArrayList<>();
        result.add(zeroLosses);
        result.add(oneLoss);

        return result;
    }
}

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Problem Info

Description

Problem 2225: Find Players With Zero or One Losses - Detailed explanation and solution approaches.

Topics

Array Hash Table Sorting Counting

Companies

IndeedPalantir Technologies

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2225

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a hash map `losses_count` to store the number of losses for each player.
2
Step 2: Iterate through the `matches` list.
3
Step 3: For each match `[winner, loser]`, update the `losses_count` hash map. Increment the count for the `loser`. If the `winner` is not in the map, insert them with a count of 0.
4
Step 4: Initialize two empty lists, `zero_losses` and `one_loss`, to store players with zero and one loss respectively.
5
Step 5: Iterate through the `losses_count` hash map.
6
Step 6: For each player and their loss count, if the count is 0, add the player to the `zero_losses` list. If the count is 1, add the player to the `one_loss` list.
7
Step 7: Sort the `zero_losses` and `one_loss` lists in ascending order.
8
Step 8: Return a list containing the `zero_losses` and `one_loss` lists.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The hash map operations (insertion and lookup) take O(1) on average. The dominant time complexity comes from sorting the lists of players with zero losses and one loss, which takes O(n log n) time using efficient sorting algorithms. The space complexity is O(n) because we store the loss counts for each player in the `losses_count` hash map. Additionally, we store players with zero and one losses in separate lists, which in the worst case can contain all the players.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using Sets

Maintain two sets: one for players who have lost and another for those who have lost only once. Iterate through the matches and update the sets accordingly. Finally, derive the zero-loss players by finding those who are not in the 'lost' set.

Time: O(m + n log n), where m is number of matches and n is number of distinct playersSpace: O(n)

Trade-offs: Could potentially be faster in some specific cases depending on set implementation, especially if the number of players is significantly larger than the number of matches and you're using a fast set implementation. However, the core idea remains similar with comparable complexity.

✨

Key Insights

1Insight 1: Use a hash map (or dictionary) to efficiently count the number of losses for each player.
2Insight 2: Iterate through the hash map to identify players with zero and one losses.
3Insight 3: Sort the resulting lists of players to meet the output requirements.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: An empty `matches` list. The code handles this correctly because the loops will not execute, and empty lists will be returned.
Edge case 2: A player only appears as a winner. The code explicitly adds such players to the `losses_count` map with a value of 0.
Edge case 3: All players have more than one loss. The `zero_losses` and `one_loss` lists will be empty, which is a valid result.
Edge case 4: A large number of players (up to 10^5). The hash map data structure efficiently handles a large number of players, and the sorting algorithm used has a reasonable time complexity.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the resulting lists. The problem specifically requires the output lists to be sorted.

✗Common mistake 2: Not handling the case where a player only appears as a winner. Failing to initialize their loss count to 0 will result in them not being included in the 'zero losses' list.

✗Common mistake 3: Using inefficient data structures or algorithms. Using a list to count losses would result in a much higher time complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing game statistics to identify top-performing players (zero losses) and players who need improvement (one loss).

🏦Real-world scenario 2: Identifying customers with perfect service records (no complaints) and customers with minor issues (one complaint) for targeted support or marketing campaigns.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find Players With Zero or One Losses"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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