What is the time complexity of Elimination Game?

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

What data structures are used to solve Elimination Game?

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

What are the key insights for solving Elimination Game?

Insight 1: Instead of actually creating the array and simulating the elimination, we can track the head (the first remaining number), the step size (the gap between remaining numbers), the direction of elimination, and the number of remaining elements. Insight 2: The key to solving this problem efficiently is understanding how the head changes after each elimination round. If we are eliminating from left to right, the head always advances by the current step size. If we are eliminating from right to left, the head advances by the current step size only when the number of remaining elements is odd.

Which companies ask LeetCode 390. Elimination Game in interviews?

This problem is frequently asked at Autodesk.

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Elimination Game - LeetCode 390 Solution

Elimination Game - Complete Solution Guide

Elimination Game is LeetCode problem 390, 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 have a list arr of all integers in the range [1, n] sorted in a strictly increasing order. Apply the following algorithm on arr : Starting from left to right, remove the first number and every other number afterward until you reach the end of the list. Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers. Keep repeating the steps again, alternating left to right and right to left, until a single number

Detailed Explanation

The problem asks us to simulate an elimination process on a list of numbers from 1 to `n`. We start by eliminating every other number from left to right. Then, we switch direction and eliminate every other number from right to left. We repeat this process, alternating directions, until only one number remains. The goal is to find this last remaining number.

Solution Approach

The solution simulates the elimination process without explicitly creating the array. It maintains four variables: `head` (the first remaining number), `step` (the gap between remaining numbers), `remaining` (the number of remaining numbers), and `from_left` (a boolean indicating the direction of elimination). In each iteration, the `head` is updated based on the elimination direction and whether the number of remaining numbers is odd or even. Then, the `remaining` numbers are halved, the `step` size is doubled, and the direction is reversed. This process continues until only one number remains.

Step-by-Step Algorithm

Step 1: Initialize `head` to 1, `step` to 1, `remaining` to `n`, and `from_left` to `True`.
Step 2: Enter a `while` loop that continues as long as `remaining` is greater than 1.
Step 3: Inside the loop, check if we are eliminating from left to right (`from_left` is `True`) or if `remaining` is odd. If either condition is true, update `head` by adding `step` to it.
Step 4: Divide `remaining` by 2 (integer division), since we eliminate half of the numbers in each iteration.
Step 5: Multiply `step` by 2, since the gap between remaining numbers doubles in each iteration.
Step 6: Toggle the value of `from_left` to switch the elimination direction.
Step 7: After the loop terminates, return the value of `head`.

Key Insights

Insight 1: Instead of actually creating the array and simulating the elimination, we can track the head (the first remaining number), the step size (the gap between remaining numbers), the direction of elimination, and the number of remaining elements.
Insight 2: The key to solving this problem efficiently is understanding how the head changes after each elimination round. If we are eliminating from left to right, the head always advances by the current step size. If we are eliminating from right to left, the head advances by the current step size only when the number of remaining elements is odd.
Insight 3: The number of remaining elements is halved in each iteration, leading to a logarithmic time complexity. The step size doubles in each iteration, reflecting the increasing gap between remaining elements.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Recursion.

Companies

Asked at: Autodesk.

Elimination Game | LeetCode 390

Medium

Math Recursion

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

Solution Code

Available in:

class Solution {
    public int lastRemaining(int n) {
        int head = 1;
        int step = 1;
        int remaining = n;
        boolean from_left = true;

        while (remaining > 1) {
            if (from_left || remaining % 2 == 1) {
                head = head + step;
            }
            
            remaining = remaining / 2;
            step = step * 2;
            from_left = !from_left;
            
        }
        return head;
    }
}

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Problem Info

Description

Problem 390: Elimination Game - Detailed explanation and solution approaches.

Topics

Math Recursion

Companies

Autodesk

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:390

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

7 steps

1
Step 1: Initialize `head` to 1, `step` to 1, `remaining` to `n`, and `from_left` to `True`.
2
Step 2: Enter a `while` loop that continues as long as `remaining` is greater than 1.
3
Step 3: Inside the loop, check if we are eliminating from left to right (`from_left` is `True`) or if `remaining` is odd. If either condition is true, update `head` by adding `step` to it.
4
Step 4: Divide `remaining` by 2 (integer division), since we eliminate half of the numbers in each iteration.
5
Step 5: Multiply `step` by 2, since the gap between remaining numbers doubles in each iteration.
6
Step 6: Toggle the value of `from_left` to switch the elimination direction.
7
Step 7: After the loop terminates, return the value of `head`.

💡

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(log(n))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The `while` loop iterates approximately log₂(n) times because the number of remaining elements is halved in each iteration. Therefore, the time complexity is O(log(n)). The algorithm uses only a few constant space variables (`head`, `step`, `remaining`, `from_left`), so the space complexity is O(1).

📖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

Recursive Approach

A recursive approach can be used to solve this problem. It's less efficient due to function call overheads, but it can provide another way to implement the same logic.

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

Trade-offs: The recursive approach is conceptually similar but less efficient due to the function call overhead and potentially increased stack usage. Iterative approach is preferred due to constant space complexity.

✨

Key Insights

1Insight 1: Instead of actually creating the array and simulating the elimination, we can track the head (the first remaining number), the step size (the gap between remaining numbers), the direction of elimination, and the number of remaining elements.
2Insight 2: The key to solving this problem efficiently is understanding how the head changes after each elimination round. If we are eliminating from left to right, the head always advances by the current step size. If we are eliminating from right to left, the head advances by the current step size only when the number of remaining elements is odd.
3Insight 3: The number of remaining elements is halved in each iteration, leading to a logarithmic time complexity. The step size doubles in each iteration, reflecting the increasing gap between remaining elements.

📚

Key Terms Explained

For Students

Recursion

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

Edge Cases

Edge case 1: n = 1. The `while` loop is never entered, and the initial value of `head` (1) is returned, which is the correct answer.
Edge case 2: n is a large number (up to 10^9). The algorithm still works correctly because it doesn't explicitly create the list of numbers. It simulates the elimination process efficiently using the `head`, `step`, `remaining`, and `from_left` variables.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly updating the `head` variable. The `head` only needs to be updated when eliminating from left to right or when eliminating from right to left and there is an odd number of elements remaining.

✗Common mistake 2: Failing to update the `step` variable correctly. The `step` must be doubled in each iteration to reflect the increasing gap between remaining numbers.

✗Common mistake 3: Trying to simulate the list directly. This leads to time limit exceeded for large values of 'n'. The efficient way is to simulate the process by keeping track of the head and the step.

✅

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 rounds where resources are allocated to every other recipient, alternating between starting from the beginning and the end.

🏦Real-world scenario 2: Data compression algorithms where data is filtered iteratively, removing elements based on alternating patterns to reduce data size.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 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: July 1, 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

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Elimination Game"?

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

🤖 Need help with this problem?