What is the time complexity of Find the Winner of an Array Game?

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

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

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

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

Insight 1: If 'k' is greater than or equal to the length of the array, the largest element in the array will win. This is because eventually the largest element will face every other element and win, before k is reached Insight 2: The problem can be solved by simulating the game. We keep track of the current winner and their consecutive win count. In each round, we compare the current winner with the next element in the array.

Which companies ask LeetCode 1535. Find the Winner of an Array Game in interviews?

This problem is frequently asked at Directi.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Find the Winner of an Array Game - LeetCode 1535 Solution

Find the Winner of an Array Game - Complete Solution Guide

Find the Winner of an Array Game is LeetCode problem 1535, 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

Given an integer array arr of distinct integers and an integer k . A game will be played between the first two elements of the array (i.e. arr[0] and arr[1] ). In each round of the game, we compare arr[0] with arr[1] , the larger integer wins and remains at position 0 , and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds. Return the integer which will win the game . It is guaranteed that there will be a winner of the game. Example 1: Inp

Detailed Explanation

The problem describes a game played between the first two elements of an array. In each round, the larger element wins and stays at the beginning, while the smaller element moves to the end of the array. The goal is to find the integer that wins 'k' consecutive rounds. The input is an array 'arr' of distinct integers and an integer 'k'. The output is the winning integer.

Solution Approach

The solution simulates the game as described in the problem statement. It maintains the current winner and a win counter. It iterates through the array, comparing the current winner with the next element. If the current winner is larger, the win counter is incremented. Otherwise, the next element becomes the new winner, and the win counter is reset to 1. The simulation stops when the win counter reaches 'k', or the end of the array is reached. If 'k' is greater or equal to the array length, we can optimize by simply returning the maximum element

Step-by-Step Algorithm

Step 1: Check if 'k' is greater than or equal to the array length. If so, find and return the maximum element in the array.
Step 2: Initialize the 'current_winner' to the first element of the array (arr[0]) and the 'win_count' to 0.
Step 3: Iterate through the array from the second element (index 1) to the end.
Step 4: In each iteration, compare the 'current_winner' with the current element (arr[i]).
Step 5: If 'current_winner' is greater than 'arr[i]', increment 'win_count'.
Step 6: Otherwise, update 'current_winner' to 'arr[i]' and reset 'win_count' to 1.
Step 7: After each comparison, check if 'win_count' is equal to 'k'. If so, return 'current_winner'.
Step 8: If the loop completes without finding a winner (win_count == k), return the 'current_winner' (as it will be the winner from the last comparison onwards).

Key Insights

Insight 1: If 'k' is greater than or equal to the length of the array, the largest element in the array will win. This is because eventually the largest element will face every other element and win, before k is reached
Insight 2: The problem can be solved by simulating the game. We keep track of the current winner and their consecutive win count. In each round, we compare the current winner with the next element in the array.
Insight 3: The problem's constraints guarantee that a winner exists, simplifying the termination condition of the simulation.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Simulation.

Companies

Asked at: Directi.

Find the Winner of an Array Game | LeetCode 1535

Medium

Array Simulation

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

Solution Code

Available in:

class Solution {
    public int getWinner(int[] arr, int k) {
        if (k >= arr.length) {
            int max = arr[0];
            for (int i = 1; i < arr.length; i++) {
                if (arr[i] > max) {
                    max = arr[i];
                }
            }
            return max;
        }

        int current_winner = arr[0];
        int win_count = 0;

        for (int i = 1; i < arr.length; i++) {
            if (current_winner > arr[i]) {
                win_count += 1;
            } else {
                current_winner = arr[i];
                win_count = 1;
            }

            if (win_count == k) {
                return current_winner;
            }
        }

        return current_winner;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1535: Find the Winner of an Array Game - Detailed explanation and solution approaches.

Topics

Array Simulation

Companies

Directi

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1535

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: Check if 'k' is greater than or equal to the array length. If so, find and return the maximum element in the array.
2
Step 2: Initialize the 'current_winner' to the first element of the array (arr[0]) and the 'win_count' to 0.
3
Step 3: Iterate through the array from the second element (index 1) to the end.
4
Step 4: In each iteration, compare the 'current_winner' with the current element (arr[i]).
5
Step 5: If 'current_winner' is greater than 'arr[i]', increment 'win_count'.
6
Step 6: Otherwise, update 'current_winner' to 'arr[i]' and reset 'win_count' to 1.
7
Step 7: After each comparison, check if 'win_count' is equal to 'k'. If so, return 'current_winner'.
8
Step 8: If the loop completes without finding a winner (win_count == k), return the 'current_winner' (as it will be the winner from the last comparison onwards).

💡

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 time complexity is O(n) because we iterate through the array at most once. The space complexity is O(1) because we only use a fixed number of variables to store the current winner, win count, and loop index.

📖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 a Deque

A Deque can be used to simulate the circular shifting of elements in the array. The winner remains at the front, and the loser is added to the end.

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

Trade-offs: This approach may be conceptually clearer for some but uses O(n) space, while the main solution uses O(1) space. Prefer the main solution for its space efficiency.

✨

Key Insights

1Insight 1: If 'k' is greater than or equal to the length of the array, the largest element in the array will win. This is because eventually the largest element will face every other element and win, before k is reached
2Insight 2: The problem can be solved by simulating the game. We keep track of the current winner and their consecutive win count. In each round, we compare the current winner with the next element in the array.
3Insight 3: The problem's constraints guarantee that a winner exists, simplifying the termination condition of the simulation.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: k >= len(arr). The solution correctly handles this by returning the maximum element, as it's guaranteed to win all subsequent rounds.
Edge case 2: Array with only two elements. The solution correctly handles this case, as the loop still executes, and the winner is determined based on their first comparison and win count.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case where 'k' is greater than or equal to the array length correctly. Forgetting this results in unnecessary iterations, though the result might still be correct it's less efficient.

✗Common mistake 2: Incorrectly updating the 'win_count' or 'current_winner', leading to an inaccurate simulation of the game.

✅

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: This algorithm can be applied to ranking systems where entities compete in rounds, and the entity with consecutive wins is declared the overall winner.

🏦Real-world scenario 2: In competitive programming contests or online gaming leaderboards, a similar concept of consecutive wins could be used to determine temporary winners or streaks.

💡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: September 4, 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 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 an Array 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 simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?