What is the time complexity of Ugly Number III?

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

What data structures are used to solve Ugly Number III?

This problem uses Math, Binary Search, Combinatorics, Number Theory. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Ugly Number III?

Insight 1: The key idea is to use binary search. Since we're looking for the nth ugly number within a certain range (1 to 2 * 10^9), we can efficiently narrow down the search space. Insight 2: The principle of inclusion-exclusion is crucial for counting the number of ugly numbers less than or equal to a given number 'mid'. We need to account for numbers divisible by a, b, and c individually, then subtract overlaps (divisible by lcm(a,b), lcm(a,c), lcm(b,c)), and finally add back the overlap of overlaps (divisible by lcm(a,b,c)).

Which companies ask LeetCode 1201. Ugly Number III in interviews?

This problem is frequently asked at American Express.

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Ugly Number III - LeetCode 1201 Solution

Ugly Number III - Complete Solution Guide

Ugly Number III is LeetCode problem 1201, 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 that is divisible by a , b , or c . Given four integers n , a , b , and c , return the n th ugly number . Example 1: Input: n = 3, a = 2, b = 3, c = 5 Output: 4 Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3 rd is 4. Example 2: Input: n = 4, a = 2, b = 3, c = 4 Output: 6 Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4 th is 6. Example 3: Input: n = 5, a = 2, b = 11, c = 13 Output: 10 Explanation: The ugly numbers are 2,

Detailed Explanation

The problem asks us to find the nth 'ugly number'. An ugly number is defined as a positive integer that is divisible by at least one of the given integers a, b, or c. The input consists of four integers: n (the desired ugly number's rank), a, b, and c (the divisors). The output should be the nth ugly number.

Solution Approach

The solution employs binary search to find the nth ugly number. For a given mid-point value, it determines how many numbers less than or equal to mid are divisible by a, b, or c. This count is obtained using the inclusion-exclusion principle. If the count is less than n, we search in the higher half; otherwise, we search in the lower half. The binary search continues until low and high converge on the nth ugly number.

Step-by-Step Algorithm

Step 1: Define a function gcd(x, y) to calculate the greatest common divisor of two numbers x and y using the Euclidean algorithm.
Step 2: Define a function lcm(x, y) to calculate the least common multiple of two numbers x and y using the formula lcm(x, y) = (x * y) / gcd(x, y).
Step 3: Calculate lcm(a, b), lcm(a, c), lcm(b, c), and lcm(a, lcm(b,c)). These values are used in the inclusion-exclusion principle.
Step 4: Initialize low to 1 and high to 2 * 10^9 (based on the problem's constraint).
Step 5: Perform binary search: While low < high, calculate mid = low + (high - low) // 2.
Step 6: Calculate the count of ugly numbers less than or equal to mid using the inclusion-exclusion principle: count = (mid // a) + (mid // b) + (mid // c) - (mid // lcm_ab) - (mid // lcm_ac) - (mid // lcm_bc) + (mid // lcm_abc).
Step 7: If count < n, it means the nth ugly number is greater than mid, so update low = mid + 1. Otherwise, the nth ugly number is less than or equal to mid, so update high = mid.
Step 8: After the loop terminates (low == high), low (or high) will be the nth ugly number. Return low.

Key Insights

Insight 1: The key idea is to use binary search. Since we're looking for the nth ugly number within a certain range (1 to 2 * 10^9), we can efficiently narrow down the search space.
Insight 2: The principle of inclusion-exclusion is crucial for counting the number of ugly numbers less than or equal to a given number 'mid'. We need to account for numbers divisible by a, b, and c individually, then subtract overlaps (divisible by lcm(a,b), lcm(a,c), lcm(b,c)), and finally add back the overlap of overlaps (divisible by lcm(a,b,c)).
Insight 3: The least common multiple (LCM) calculations can result in large numbers, so using long data types in Java/C++ and proper integer division in Python is important to avoid overflow.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Binary Search, Combinatorics, Number Theory.

Companies

Asked at: American Express.

Ugly Number III | LeetCode 1201

Medium

Math Binary Search Combinatorics Number Theory

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

Solution Code

Available in:

class Solution {
    public int nthUglyNumber(int n, int a, int b, int c) {
        
        long lcm_ab = lcm(a, b);
        long lcm_ac = lcm(a, c);
        long lcm_bc = lcm(b, c);
        long lcm_abc = lcm(a, lcm_bc);

        long low = 1;
        long high = (long)2e9;

        while (low < high) {
            long mid = low + (high - low) / 2;
            
            long count = (mid / a) + (mid / b) + (mid / c)
                  - (mid / lcm_ab) - (mid / lcm_ac) - (mid / lcm_bc)
                  + (mid / lcm_abc);
            
            if (count < n) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
                
        return (int)low;
    }
    
    private long gcd(long x, long y) {
            while (y != 0) {
                long temp = y;
                y = x % y;
                x = temp;
            }
            return x;
        }

    private long lcm(long x, long y) {
        return (x * y) / gcd(x, y);
    }
}

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Problem Info

Description

Problem 1201: Ugly Number III - Detailed explanation and solution approaches.

Topics

Math Binary Search Combinatorics Number Theory

Companies

American Express

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1201

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: Define a function gcd(x, y) to calculate the greatest common divisor of two numbers x and y using the Euclidean algorithm.
2
Step 2: Define a function lcm(x, y) to calculate the least common multiple of two numbers x and y using the formula lcm(x, y) = (x * y) / gcd(x, y).
3
Step 3: Calculate lcm(a, b), lcm(a, c), lcm(b, c), and lcm(a, lcm(b,c)). These values are used in the inclusion-exclusion principle.
4
Step 4: Initialize low to 1 and high to 2 * 10^9 (based on the problem's constraint).
5
Step 5: Perform binary search: While low < high, calculate mid = low + (high - low) // 2.
6
Step 6: Calculate the count of ugly numbers less than or equal to mid using the inclusion-exclusion principle: count = (mid // a) + (mid // b) + (mid // c) - (mid // lcm_ab) - (mid // lcm_ac) - (mid // lcm_bc) + (mid // lcm_abc).
7
Step 7: If count < n, it means the nth ugly number is greater than mid, so update low = mid + 1. Otherwise, the nth ugly number is less than or equal to mid, so update high = mid.
8
Step 8: After the loop terminates (low == high), low (or high) will be the nth ugly number. Return low.

💡

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?

Binary search divides the search space in half at each step. The LCM and GCD computations take constant time. Therefore, the dominant factor in terms of time complexity is the binary search, which is O(log(range)). Since the range is fixed at 2*10^9, the complexity is effectively O(log(2*10^9)), which can be considered O(log(n)) relative to a possible large value of n given in a general sense (though not strictly linked to the input 'n' of the problem itself). The space complexity is constant because only a few variables are used, regardless of the size of the input.

📖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-based approach

Use a min-heap to maintain a set of candidate ugly numbers. Initialize the heap with a, b, and c. Repeatedly extract the smallest element from the heap, multiply it by a, b, and c, and add the results to the heap (if they are not already present). This approach is similar to finding the nth super ugly number.

Time: O(n log n) in the worst caseSpace: O(n)

Trade-offs: The heap approach is generally slower than binary search (O(n log n) vs O(log n)), especially for larger values of n. It also requires more space to store the heap. However, it avoids the need for LCM calculations and inclusion-exclusion.

✨

Key Insights

1Insight 1: The key idea is to use binary search. Since we're looking for the nth ugly number within a certain range (1 to 2 * 10^9), we can efficiently narrow down the search space.
2Insight 2: The principle of inclusion-exclusion is crucial for counting the number of ugly numbers less than or equal to a given number 'mid'. We need to account for numbers divisible by a, b, and c individually, then subtract overlaps (divisible by lcm(a,b), lcm(a,c), lcm(b,c)), and finally add back the overlap of overlaps (divisible by lcm(a,b,c)).
3Insight 3: The least common multiple (LCM) calculations can result in large numbers, so using long data types in Java/C++ and proper integer division in Python is important to avoid overflow.

📚

Key Terms Explained

For Students

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: a, b, or c being equal to 1. This will result in the first ugly number being 1, and the algorithm correctly handles it due to the inclusion-exclusion principle and binary search.
Edge case 2: Very large values of a, b, or c approaching 10^9. This is handled by using long data types to prevent integer overflow in the LCM calculations.
Edge case 3: Duplicate values among a, b, and c. The LCM and inclusion-exclusion principle correctly account for this duplication.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow in LCM calculations. This can lead to incorrect results. Using `long` data type (or equivalent in other languages) is crucial to prevent this.

✗Common mistake 2: Not using the inclusion-exclusion principle correctly. Failing to account for overlaps between multiples of a, b, and c will result in an incorrect count of ugly numbers.

✗Common mistake 3: Incorrect binary search implementation. It's important to correctly update low and high based on whether the count is less than n or not. Off-by-one errors are common.

✅

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 systems might use a similar concept to generate unique identifiers or primary keys based on a set of predefined rules.

🏦Real-world scenario 2: Compiler optimization techniques can utilize number theory concepts (like divisibility) to improve code efficiency.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: September 30, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

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 "Ugly Number III"?

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

🤖 Need help with this problem?