What is the time complexity of Ugly Number II?

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

What data structures are used to solve Ugly Number II?

This problem uses Hash Table, Math, Dynamic Programming, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Ugly Number II?

Insight 1: Ugly numbers can be generated by multiplying existing ugly numbers by 2, 3, or 5. Insight 2: Dynamic programming can be used to store and build the sequence of ugly numbers efficiently.

Which companies ask LeetCode 264. Ugly Number II in interviews?

This problem is frequently asked at Accenture, EPAM Systems.

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Ugly Number II - LeetCode 264 Solution

Q: How to solve Ugly Number II?

The solution uses dynamic programming to generate the sequence of ugly numbers. It maintains an array `ugly` to store the ugly numbers in ascending order. The algorithm iteratively calculates the next ugly number by finding the minimum of the products of the existing ugly numbers with 2, 3, and 5. Three pointers, `p2`, `p3`, and `p5`, are used to track the indices of the ugly numbers that will be multiplied by 2, 3, and 5, respectively. After finding the minimum (which is the next ugly number), the corresponding pointer(s) are incremented. If a number can be generated in multiple ways, each corresponding pointer is incremented to avoid duplicates.

Ugly Number II - Complete Solution Guide

Ugly Number II is LeetCode problem 264, 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

An ugly number is a positive integer whose prime factors are limited to 2 , 3 , and 5 . Given an integer n , return the n th ugly number . Example 1: Input: n = 10 Output: 12 Explanation: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12] is the sequence of the first 10 ugly numbers. Example 2: Input: n = 1 Output: 1 Explanation: 1 has no prime factors, therefore all of its prime factors are limited to 2, 3, and 5. Constraints: 1 <= n <= 1690

Detailed Explanation

The problem asks us to find the n-th ugly number. An ugly number is a positive integer whose prime factors are limited to 2, 3, and 5. For example, the first 10 ugly numbers are [1, 2, 3, 4, 5, 6, 8, 9, 10, 12]. The input is an integer 'n' representing the position of the ugly number we want to find (1-indexed). The output is the n-th ugly number.

Solution Approach

The solution uses dynamic programming to generate the sequence of ugly numbers. It maintains an array `ugly` to store the ugly numbers in ascending order. The algorithm iteratively calculates the next ugly number by finding the minimum of the products of the existing ugly numbers with 2, 3, and 5. Three pointers, `p2`, `p3`, and `p5`, are used to track the indices of the ugly numbers that will be multiplied by 2, 3, and 5, respectively. After finding the minimum (which is the next ugly number), the corresponding pointer(s) are incremented. If a number can be generated in multiple ways, each corresponding pointer is incremented to avoid duplicates.

Step-by-Step Algorithm

Step 1: Initialize an array `ugly` of size `n` to store the ugly numbers. Initialize `ugly[0]` to 1, as 1 is the first ugly number.
Step 2: Initialize three pointers `p2`, `p3`, and `p5` to 0. These pointers track the indices of the ugly numbers that will be multiplied by 2, 3, and 5, respectively.
Step 3: Iterate from `i = 1` to `n - 1` to calculate the remaining ugly numbers.
Step 4: Calculate `c2 = ugly[p2] * 2`, `c3 = ugly[p3] * 3`, and `c5 = ugly[p5] * 5`. These are the candidate ugly numbers.
Step 5: Find the minimum of `c2`, `c3`, and `c5`. This minimum value is the next ugly number `ugly[i]`.
Step 6: Increment the pointer(s) corresponding to the factor(s) that produced the minimum value. This is crucial to avoid generating duplicates. For instance, if `c2` and `c3` are both equal to the minimum, increment both `p2` and `p3`.
Step 7: After the loop, the n-th ugly number is stored in `ugly[n - 1]`.

Key Insights

Insight 1: Ugly numbers can be generated by multiplying existing ugly numbers by 2, 3, or 5.
Insight 2: Dynamic programming can be used to store and build the sequence of ugly numbers efficiently.
Insight 3: Maintaining separate pointers for each prime factor (2, 3, 5) allows us to generate the next ugly number in increasing order without redundant calculations or sorting.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Math, Dynamic Programming, Heap (Priority Queue).

Companies

Asked at: Accenture, EPAM Systems.

Ugly Number II | LeetCode 264

Medium

Hash Table Math Dynamic Programming Heap (Priority Queue)

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

Solution Code

Available in:

class Solution {
    public int nthUglyNumber(int n) {
        int[] ugly = new int[n];
        ugly[0] = 1;
        int p2 = 0, p3 = 0, p5 = 0;

        for (int i = 1; i < n; i++) {
            int c2 = ugly[p2] * 2;
            int c3 = ugly[p3] * 3;
            int c5 = ugly[p5] * 5;

            int min_val = Math.min(c2, Math.min(c3, c5));
            ugly[i] = min_val;

            if (min_val == c2) {
                p2 += 1;
            }
            if (min_val == c3) {
                p3 += 1;
            }
            if (min_val == c5) {
                p5 += 1;
            }
        }

        return ugly[n - 1];
    }
}

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Problem Info

Description

Problem 264: Ugly Number II - Detailed explanation and solution approaches.

Topics

Hash Table Math Dynamic Programming Heap (Priority Queue)

Companies

AccentureEPAM Systems

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:264

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 an array `ugly` of size `n` to store the ugly numbers. Initialize `ugly[0]` to 1, as 1 is the first ugly number.
2
Step 2: Initialize three pointers `p2`, `p3`, and `p5` to 0. These pointers track the indices of the ugly numbers that will be multiplied by 2, 3, and 5, respectively.
3
Step 3: Iterate from `i = 1` to `n - 1` to calculate the remaining ugly numbers.
4
Step 4: Calculate `c2 = ugly[p2] * 2`, `c3 = ugly[p3] * 3`, and `c5 = ugly[p5] * 5`. These are the candidate ugly numbers.
5
Step 5: Find the minimum of `c2`, `c3`, and `c5`. This minimum value is the next ugly number `ugly[i]`.
6
Step 6: Increment the pointer(s) corresponding to the factor(s) that produced the minimum value. This is crucial to avoid generating duplicates. For instance, if `c2` and `c3` are both equal to the minimum, increment both `p2` and `p3`.
7
Step 7: After the loop, the n-th ugly number is stored in `ugly[n - 1]`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because the algorithm iterates 'n' times to generate the ugly numbers. The space complexity is O(n) because we are using an array of size 'n' to store the ugly numbers.

📖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

Heap (Priority Queue)

Another approach is to use a min-heap (priority queue) to store the ugly numbers. We start with 1 in the heap. Then, repeatedly remove the smallest element from the heap, and add its multiples (2, 3, and 5) to the heap, ensuring that duplicates are not added using a set.

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

Trade-offs: The heap approach has a time complexity of O(n log n) due to the heap operations, making it slightly less efficient than the dynamic programming approach, especially when n is relatively small. However, it might be useful when you only need to calculate specific ugly numbers and not the entire sequence or when the problem constraints change to require more complex prime factors.

✨

Key Insights

1Insight 1: Ugly numbers can be generated by multiplying existing ugly numbers by 2, 3, or 5.
2Insight 2: Dynamic programming can be used to store and build the sequence of ugly numbers efficiently.
3Insight 3: Maintaining separate pointers for each prime factor (2, 3, 5) allows us to generate the next ugly number in increasing order without redundant calculations or sorting.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The solution correctly returns 1 as the first ugly number.
Edge case 2: n is a large number close to the constraint limit (1690). The solution correctly generates the sequence up to that limit.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to increment all pointers when multiple factors produce the same minimum value. This can lead to missing ugly numbers in the sequence.

✗Common mistake 2: Not using `long` data type for intermediate calculations in C++ or C. Multiplying larger ugly numbers by 2, 3, or 5 can cause integer overflow, leading to incorrect results. Python handles this implicitly.

✗Common mistake 3: Incorrectly implementing the min-heap approach leading to duplicate entries or higher 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: Database indexing where the selection of certain key values are based on optimized prime factorization. Specifically, creating identifiers or keys based on multiplying smaller set of prime numbers.

🏦Real-world scenario 2: In some resource allocation problems, resources might be allocated based on 'ugly' values to ensure fairness and avoid biases inherent in consecutive integer assignments. This can be seen in queuing systems or task scheduling.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

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

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate 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 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 "Ugly Number II"?

For this medium-level Hash Table 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.

🤖 Need help with this problem?