What is the time complexity of Ugly Number?

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

What data structures are used to solve Ugly Number?

This problem uses Math. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Ugly Number?

Insight 1: Repeatedly dividing by 2, 3, and 5: The core idea is to repeatedly divide the input number by 2, 3, and 5 until it's no longer divisible by any of them. If the final result is 1, it means the original number only had 2, 3, and 5 as prime factors. Insight 2: Handling negative and zero inputs: The problem explicitly states that an ugly number is a *positive* integer. Therefore, negative numbers and zero are not ugly numbers and should be handled separately.

Which companies ask LeetCode 263. Ugly Number in interviews?

This problem is frequently asked at J.P. Morgan.

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

Ugly Number - LeetCode 263 Solution

Q: How to solve Ugly Number?

The solution uses a straightforward approach. It first checks if the input `n` is less than or equal to 0. If so, it immediately returns `false` because ugly numbers must be positive. Otherwise, it iteratively divides `n` by 2, 3, and 5 as long as it's divisible. After this process, if `n` is equal to 1, it means that all prime factors were 2, 3, or 5, thus the number is ugly; otherwise, it's not. This is done using `while` loops for efficiency.

Ugly Number - Complete Solution Guide

Ugly Number is LeetCode problem 263, 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

An ugly number is a positive integer which does not have a prime factor other than 2, 3, and 5. Given an integer n , return true if n is an ugly number . Example 1: Input: n = 6 Output: true Explanation: 6 = 2 × 3 Example 2: Input: n = 1 Output: true Explanation: 1 has no prime factors. Example 3: Input: n = 14 Output: false Explanation: 14 is not ugly since it includes the prime factor 7. Constraints: -2 31 <= n <= 2 31 - 1

Detailed Explanation

The problem asks you to determine if a given integer `n` is an 'ugly number'. An ugly number is a positive integer whose only prime factors are 2, 3, or 5. The input is an integer `n`, and the output is a boolean value: `true` if `n` is an ugly number, and `false` otherwise. The input integer can be positive, negative, or zero.

Solution Approach

The solution uses a straightforward approach. It first checks if the input `n` is less than or equal to 0. If so, it immediately returns `false` because ugly numbers must be positive. Otherwise, it iteratively divides `n` by 2, 3, and 5 as long as it's divisible. After this process, if `n` is equal to 1, it means that all prime factors were 2, 3, or 5, thus the number is ugly; otherwise, it's not. This is done using `while` loops for efficiency.

Step-by-Step Algorithm

Step 1: Check if `n` is less than or equal to 0. If true, return `false`.
Step 2: Divide `n` by 2 repeatedly until it's no longer divisible by 2.
Step 3: Divide `n` by 3 repeatedly until it's no longer divisible by 3.
Step 4: Divide `n` by 5 repeatedly until it's no longer divisible by 5.
Step 5: Check if `n` is equal to 1. If true, return `true`; otherwise, return `false`.

Key Insights

Insight 1: Repeatedly dividing by 2, 3, and 5: The core idea is to repeatedly divide the input number by 2, 3, and 5 until it's no longer divisible by any of them. If the final result is 1, it means the original number only had 2, 3, and 5 as prime factors.
Insight 2: Handling negative and zero inputs: The problem explicitly states that an ugly number is a *positive* integer. Therefore, negative numbers and zero are not ugly numbers and should be handled separately.
Insight 3: Efficiency of the while loops: The `while` loops are efficient because they continue dividing until the number is no longer divisible by the prime factor. This avoids unnecessary iterations.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Math.

Companies

Asked at: J.P. Morgan.

Ugly Number | LeetCode 263

Easy

Math

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

Solution Code

Available in:

class Solution {
    public boolean isUgly(int n) {
        if (n <= 0) return false;
        while (n % 2 == 0) n /= 2;
        while (n % 3 == 0) n /= 3;
        while (n % 5 == 0) n /= 5;
        return n == 1;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 263: Ugly Number - Detailed explanation and solution approaches.

Topics

Math

Companies

J.P. Morgan

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:263

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: Check if `n` is less than or equal to 0. If true, return `false`.
2
Step 2: Divide `n` by 2 repeatedly until it's no longer divisible by 2.
3
Step 3: Divide `n` by 3 repeatedly until it's no longer divisible by 3.
4
Step 4: Divide `n` by 5 repeatedly until it's no longer divisible by 5.
5
Step 5: Check if `n` is equal to 1. If true, return `true`; otherwise, return `false`.

💡

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 time complexity is logarithmic because in the worst case, the number of times we divide `n` by 2, 3, or 5 is proportional to the logarithm of `n` (base 2, 3, or 5). The space complexity is constant because we are only using a few integer variables.

📖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

Prime Factorization

Find all prime factors of `n`. Then check if all prime factors are 2, 3, or 5.

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

Trade-offs: This approach might be slightly less efficient than the provided solution for large numbers as it involves finding all prime factors. However, it provides a clearer way to understand the concept of prime factorization in relation to ugly numbers.

✨

Key Insights

1Insight 1: Repeatedly dividing by 2, 3, and 5: The core idea is to repeatedly divide the input number by 2, 3, and 5 until it's no longer divisible by any of them. If the final result is 1, it means the original number only had 2, 3, and 5 as prime factors.
2Insight 2: Handling negative and zero inputs: The problem explicitly states that an ugly number is a *positive* integer. Therefore, negative numbers and zero are not ugly numbers and should be handled separately.
3Insight 3: Efficiency of the while loops: The `while` loops are efficient because they continue dividing until the number is no longer divisible by the prime factor. This avoids unnecessary iterations.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n = 0 or n < 0: The code correctly handles these cases by returning `false` immediately.
Edge case 2: n = 1: The code correctly identifies 1 as an ugly number because it has no prime factors other than 2, 3, and 5 (or none at all).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle negative or zero inputs: Ugly numbers are defined as positive integers. Failing to check for non-positive inputs will result in incorrect results.

✗Common mistake 2: Using inefficient division: Instead of repeatedly dividing until the remainder is 0 (using a `while` loop), inefficiently checking divisibility multiple times in a loop would increase 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: Number theory problems: This algorithm is relevant in various number theory problems involving prime factorization and divisibility.

🏦Real-world scenario 2: Data validation and filtering: You might need to check if numbers in a dataset are ugly numbers, for example in a system where only numbers with specific prime factors are allowed.

💡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.

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 23, 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 13

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

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

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 "Ugly Number"?

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

🤖 Need help with this problem?