What is the time complexity of Find the Winner of the Circular Game?

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

What data structures are used to solve Find the Winner of the Circular Game?

This problem uses Array, Math, Recursion, Queue, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find the Winner of the Circular Game?

Insight 1: The problem can be solved without explicitly simulating the circle. Instead, we can use modular arithmetic to keep track of the position of the winner. Insight 2: The key is to work backwards. The winner of the game with `n` people is related to the winner of the game with `n-1` people.

Which companies ask LeetCode 1823. Find the Winner of the Circular Game in interviews?

This problem is frequently asked at Accenture, Arista Networks, Groupon, SoFi, Zoho.

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Find the Winner of the Circular Game - LeetCode 1823 Solution

Q: How to solve Find the Winner of the Circular Game?

The provided solution employs a mathematical approach based on modular arithmetic. It iterates from 1 to `n`, simulating the elimination process in reverse. The `winner_pos` variable keeps track of the winner's position in the shrinking circle. In each iteration, we update the `winner_pos` using the formula `(winner_pos + k) % i`, where `i` represents the current number of remaining friends. Finally, we add 1 to the resulting `winner_pos` because the friend numbers start from 1, not 0.

Find the Winner of the Circular Game - Complete Solution Guide

Find the Winner of the Circular Game is LeetCode problem 1823, 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

There are n friends that are playing a game. The friends are sitting in a circle and are numbered from 1 to n in clockwise order . More formally, moving clockwise from the i th friend brings you to the (i+1) th friend for 1 <= i < n , and moving clockwise from the n th friend brings you to the 1 st friend. The rules of the game are as follows: Start at the 1 st friend. Count the next k friends in the clockwise direction including the friend you started at. The counting wraps around the circle an

Detailed Explanation

The problem describes a circular game where `n` friends are arranged in a circle, and every `k`th friend is eliminated until only one winner remains. The goal is to determine which friend wins the game, starting the elimination process from friend number 1.

Solution Approach

The provided solution employs a mathematical approach based on modular arithmetic. It iterates from 1 to `n`, simulating the elimination process in reverse. The `winner_pos` variable keeps track of the winner's position in the shrinking circle. In each iteration, we update the `winner_pos` using the formula `(winner_pos + k) % i`, where `i` represents the current number of remaining friends. Finally, we add 1 to the resulting `winner_pos` because the friend numbers start from 1, not 0.

Step-by-Step Algorithm

Step 1: Initialize `winner_pos` to 0. This variable will store the index (0-based) of the winner.
Step 2: Iterate from `i = 1` to `n`. In each iteration, `i` represents the number of remaining friends.
Step 3: Update `winner_pos` using the formula `winner_pos = (winner_pos + k) % i`. This calculates the new winner's position after eliminating one friend from the circle of size `i`.
Step 4: After the loop finishes, add 1 to `winner_pos` to convert it to the 1-based indexing required by the problem. This gives the final winner number.
Step 5: Return the final winner number.

Key Insights

Insight 1: The problem can be solved without explicitly simulating the circle. Instead, we can use modular arithmetic to keep track of the position of the winner.
Insight 2: The key is to work backwards. The winner of the game with `n` people is related to the winner of the game with `n-1` people.
Insight 3: The problem asks for a solution with linear time and constant space, suggesting that a direct simulation (using a linked list or array) might not be the most efficient approach.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Recursion, Queue, Simulation.

Companies

Asked at: Accenture, Arista Networks, Groupon, SoFi, Zoho.

Find the Winner of the Circular Game | LeetCode 1823

Medium

Array Math Recursion Queue Simulation

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int findTheWinner(int n, int k) {
        int winner_pos = 0;
        for (int i = 1; i <= n; i++) {
            winner_pos = (winner_pos + k) % i;
        }
        return winner_pos + 1;
    }
}

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Problem Info

Description

Problem 1823: Find the Winner of the Circular Game - Detailed explanation and solution approaches.

Topics

Array Math Recursion Queue Simulation

Companies

AccentureArista NetworksGrouponSoFiZoho

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:1823

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

5 steps

1
Step 1: Initialize `winner_pos` to 0. This variable will store the index (0-based) of the winner.
2
Step 2: Iterate from `i = 1` to `n`. In each iteration, `i` represents the number of remaining friends.
3
Step 3: Update `winner_pos` using the formula `winner_pos = (winner_pos + k) % i`. This calculates the new winner's position after eliminating one friend from the circle of size `i`.
4
Step 4: After the loop finishes, add 1 to `winner_pos` to convert it to the 1-based indexing required by the problem. This gives the final winner number.
5
Step 5: Return the final winner number.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution iterates from 1 to `n` in a single loop, making the time complexity O(n). The solution uses only a constant amount of extra space for variables like `winner_pos`, regardless of the input size, hence O(1) space complexity.

📖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

Simulation with Circular Linked List or Array

Simulate the game directly using a circular linked list or an array. Remove the `k`th element in each round until only one element remains.

Time: O(n*k) in the worst case. Although each removal takes O(1) with a linked list, finding the k-th element still takes O(k). And we have n removal rounds.Space: O(n) to store the list or array.

Trade-offs: This approach is more intuitive but less efficient, especially for larger values of `n` and `k`. The O(n) space complexity also makes it less desirable than the provided solution.

Recursion

Solve recursively. Let `winner(n, k)` be the winner of the game with `n` friends and step size `k`. Then `winner(n, k) = (winner(n-1, k) + k) % n`. The base case is `winner(1, k) = 0`.

Time: O(n)Space: O(n) due to recursion stack

Trade-offs: The recursion solution is functionally similar to the iterative solution. However, it has an O(n) space complexity due to the recursive call stack, making the iterative solution more space-efficient.

✨

Key Insights

1Insight 1: The problem can be solved without explicitly simulating the circle. Instead, we can use modular arithmetic to keep track of the position of the winner.
2Insight 2: The key is to work backwards. The winner of the game with `n` people is related to the winner of the game with `n-1` people.
3Insight 3: The problem asks for a solution with linear time and constant space, suggesting that a direct simulation (using a linked list or array) might not be the most efficient approach.

📚

Key Terms Explained

For Students

Array

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

Recursion

A function that calls itself to solve smaller versions of the same problem.

Queue

FIFO structure - First In, First Out. Like a line at a store!

Edge Cases

Edge case 1: `n = 1`. The loop runs only once, and `winner_pos` becomes `(0 + k) % 1 = 0`. Adding 1, the function returns 1, which is the correct winner when there's only one friend.
Edge case 2: `k = 1`. This means the friend immediately clockwise is eliminated each time. The solution correctly handles this, updating the `winner_pos` appropriately in each iteration.
Edge case 3: `k = n`. When k is equal to n, the last person is eliminated in the first round, then the problem reduces to a smaller circle.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to add 1 to `winner_pos` at the end. The result represents the 0-based index of the winner, but the problem requires the 1-based friend number.

✗Common mistake 2: Incorrectly implementing the modular arithmetic, leading to incorrect winner positions. The formula `(winner_pos + k) % i` is crucial for correctly simulating the circular elimination.

✗Common mistake 3: Trying to simulate the entire game using an array or linked list, leading to Time Limit Exceeded (TLE) errors, particularly for large values of `n` and `k`.

✅

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: Resource allocation in a circular manner. Imagine allocating resources (e.g., time slots, network bandwidth) to clients arranged in a circle. A similar algorithm could be used to determine which client receives the last available resource after repeatedly allocating to every 'k'th client.

🏦Real-world scenario 2: Load balancing in distributed systems. In a distributed system with multiple servers, tasks could be assigned in a round-robin fashion. The algorithm could be used to predict which server would handle the last task if some servers become unavailable (analogous to being eliminated).

💡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.

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: June 24, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

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 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 the Winner of the Circular Game"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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