What is the time complexity of Divisor Game?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Divisor Game?

This problem uses Math, Dynamic Programming, Brainteaser, Game Theory. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Divisor Game?

The game's outcome is solely determined by the initial number's parity: whether `n` is even or odd. This is the fundamental pattern that governs optimal play. When `n` is even (and `n > 1`), choosing `x=1` is always a valid and optimal move. This action changes the number to `n-1`, an odd number, effectively forcing the opponent into a state where they must pass an even number back.

Which companies ask LeetCode 1025. Divisor Game in interviews?

This problem is frequently asked at Visa.

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Divisor Game - LeetCode 1025 Solution

Q: How to solve Divisor Game?

The deceptively simple solution `return n % 2 == 0` reveals that the game's outcome hinges entirely on the *parity* of the initial number `n`. The key insight is that a player can always control the parity of the number they pass to their opponent if they start with an even number. If `n` is even (and `n > 1`), Alice can always choose `x=1`. Since `1` is a divisor of any integer and `0 1`, this is a valid move. By choosing `x=1`, Alice transforms `n` into `n-1`, which is an odd number. Conversely, if `n` is odd, any of its divisors `x` must also be odd. Therefore, `n-x` will always be an even number (odd - odd = even). This means a player who receives an odd number is *forced* to pass an even number back to their opponent. Since `n=1` is the losing state (and `1` is odd), the player who can consistently pass odd numbers will eventually force their opponent to receive `n=1`. Alice, starting with an even `n`, can always use the `x=1` strategy to pass an odd number to Bob, setting him up for failure. Bob, starting with an odd number, is forced to pass an even number back to Alice, putting her in a winning position. This continues until Alice passes `1` to Bob, thus winning the game.

Divisor Game - Complete Solution Guide

Divisor Game is LeetCode problem 1025, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Alice and Bob take turns playing a game, with Alice starting first. Initially, there is a number n on the chalkboard. On each player's turn, that player makes a move consisting of: Choosing any x with 0 < x < n and n % x == 0 . Replacing the number n on the chalkboard with n - x . Also, if a player cannot make a move, they lose the game. Return true if and only if Alice wins the game, assuming both players play optimally . Example 1: Input: n = 2 Output: true Explanation: Alice chooses 1, and Bo

Detailed Explanation

Alice and Bob are locked in a mathematical duel, starting with a number `n` on the chalkboard. The core challenge is understanding how their choices for `x` (a divisor of `n`, where `0 < x < n`) fundamentally alter the game state. Each turn, a player replaces `n` with `n - x`. The critical rule is that if a player cannot make a move (i.e., no such `x` exists), they lose. This immediately tells us that `n=1` is a losing state, as there are no divisors `x` satisfying `0 < x < 1`. The game revolves around optimal play: Alice wins if she can force Bob into a losing state, and Bob wins if he can force Alice into a losing state. For example, if `n=3`, Alice must choose `x=1` (the only valid divisor for `n=3`). The number becomes `3-1=2`. Bob now faces `n=2`. His only move is `x=1`, making the number `2-1=1`. Alice then receives `n=1` and loses. So, starting with `n=3`, Alice loses.

Solution Approach

The deceptively simple solution `return n % 2 == 0` reveals that the game's outcome hinges entirely on the *parity* of the initial number `n`. The key insight is that a player can always control the parity of the number they pass to their opponent if they start with an even number. If `n` is even (and `n > 1`), Alice can always choose `x=1`. Since `1` is a divisor of any integer and `0 < 1 < n` is true for `n > 1`, this is a valid move. By choosing `x=1`, Alice transforms `n` into `n-1`, which is an odd number. Conversely, if `n` is odd, any of its divisors `x` must also be odd. Therefore, `n-x` will always be an even number (odd - odd = even). This means a player who receives an odd number is *forced* to pass an even number back to their opponent. Since `n=1` is the losing state (and `1` is odd), the player who can consistently pass odd numbers will eventually force their opponent to receive `n=1`. Alice, starting with an even `n`, can always use the `x=1` strategy to pass an odd number to Bob, setting him up for failure. Bob, starting with an odd number, is forced to pass an even number back to Alice, putting her in a winning position. This continues until Alice passes `1` to Bob, thus winning the game.

Step-by-Step Algorithm

Step 1: Check if the input number `n` is even using the modulo operator (`n % 2`).
Step 2: If `n % 2` is equal to 0 (meaning `n` is even), return `true` (Alice wins).
Step 3: Otherwise (if `n` is odd), return `false` (Alice loses).

Key Insights

The game's outcome is solely determined by the initial number's parity: whether `n` is even or odd. This is the fundamental pattern that governs optimal play.
When `n` is even (and `n > 1`), choosing `x=1` is always a valid and optimal move. This action changes the number to `n-1`, an odd number, effectively forcing the opponent into a state where they must pass an even number back.
If `n` is odd, all its divisors `x` must also be odd. Consequently, any move `n-x` will result in an even number (odd minus odd equals even). This means the player starting with an odd `n` is unable to pass an odd number to their opponent, thus losing the strategic advantage of controlling parity.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming, Brainteaser, Game Theory.

Companies

Asked at: Visa.

Divisor Game | LeetCode 1025

Easy

Math Dynamic Programming Brainteaser Game Theory

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

Solution Code

Available in:

class Solution {
    public boolean divisorGame(int n) {
        return n % 2 == 0;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1025: Divisor Game - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Brainteaser Game Theory

Companies

Visa

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1025

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

3 steps

1
Step 1: Check if the input number `n` is even using the modulo operator (`n % 2`).
2
Step 2: If `n % 2` is equal to 0 (meaning `n` is even), return `true` (Alice wins).
3
Step 3: Otherwise (if `n` is odd), return `false` (Alice loses).

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution involves a single modulo operation, which takes constant time regardless of the input size. No additional data structures are used, resulting in constant 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

Recursive Approach

A recursive solution could be implemented to simulate the game. This approach would recursively explore all possible moves for each player, ultimately determining the winner.

Time: O(n*d(n)) where d(n) is the number of divisors of n. This is far less efficient than the provided solution.Space: O(n) due to the recursive call stack in the worst case.

Trade-offs: While conceptually clear, the recursive approach is significantly less efficient than the direct parity check. It is only suitable for educational purposes or smaller values of n.

✨

Key Insights

1The game's outcome is solely determined by the initial number's parity: whether `n` is even or odd. This is the fundamental pattern that governs optimal play.
2When `n` is even (and `n > 1`), choosing `x=1` is always a valid and optimal move. This action changes the number to `n-1`, an odd number, effectively forcing the opponent into a state where they must pass an even number back.
3If `n` is odd, all its divisors `x` must also be odd. Consequently, any move `n-x` will result in an even number (odd minus odd equals even). This means the player starting with an odd `n` is unable to pass an odd number to their opponent, thus losing the strategic advantage of controlling parity.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The solution correctly identifies that Alice loses (returns `false`) because she cannot make a move.
Edge case 2: n = 1000. The solution correctly identifies that Alice wins (returns `true`) because 1000 is even.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming that the game's outcome depends on other factors besides the parity of `n`. Beginners might try to analyze the divisors in more detail, leading to unnecessary complexity.

✗Common mistake 2: Not recognizing the optimal strategy. Failing to understand that choosing a divisor that leaves an odd number for the opponent is crucial to winning.

✅

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 problem demonstrates a simple instance of game theory, where understanding optimal strategies is key to winning. This concept is applicable in many areas, including competitive programming, economics, and even some aspects of cybersecurity.

🏦Real-world scenario 2: The underlying concept of parity and its significance in determining outcomes can be found in various algorithms and data structures, such as checking for even/odd indices in arrays or manipulating bits in computer systems.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Math. It's a great foundation for more complex problems.

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: October 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

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Divisor Game"?

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

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