What is the time complexity of Bulb Switcher?

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

What data structures are used to solve Bulb Switcher?

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

What are the key insights for solving Bulb Switcher?

Insight 1: A bulb will be ON at the end if it has been toggled an odd number of times. It will be OFF if toggled an even number of times. Insight 2: The number of times a bulb is toggled is equal to the number of its divisors. For example, bulb number 12 is toggled in rounds 1, 2, 3, 4, 6, and 12 (6 divisors).

Which companies ask LeetCode 319. Bulb Switcher in interviews?

This problem is frequently asked at Accenture.

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Bulb Switcher - LeetCode 319 Solution

Bulb Switcher - Complete Solution Guide

Bulb Switcher is LeetCode problem 319, 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

There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Return the number of bulbs that are on after n rounds. Example 1: Input: n = 3 Output: 1 Explanation: At first, the three bulbs are [off, off, off]. After the first round, the three bulbs are [o

Detailed Explanation

The problem asks us to simulate a series of bulb toggling rounds. Initially, all 'n' bulbs are off. In the first round, all bulbs are turned on. In the second round, every second bulb is toggled (turned off). In the third round, every third bulb is toggled, and so on. After 'n' rounds, we need to determine how many bulbs are left on.

Solution Approach

The solution leverages the mathematical property that a number has an odd number of divisors if and only if it's a perfect square. Therefore, we need to count the number of perfect squares less than or equal to 'n'. This can be done by finding the integer square root of 'n'.

Step-by-Step Algorithm

Step 1: Calculate the integer square root of 'n'. This effectively finds the largest integer 'x' such that x*x <= n.
Step 2: Return the integer square root. This value represents the count of perfect squares from 1 to 'n' (1*1, 2*2, 3*3, ... x*x).

Key Insights

Insight 1: A bulb will be ON at the end if it has been toggled an odd number of times. It will be OFF if toggled an even number of times.
Insight 2: The number of times a bulb is toggled is equal to the number of its divisors. For example, bulb number 12 is toggled in rounds 1, 2, 3, 4, 6, and 12 (6 divisors).
Insight 3: A number has an odd number of divisors if and only if it is a perfect square. For example, 9 has divisors 1, 3, and 9 (3 divisors). If a number is not a perfect square, its divisors come in pairs (e.g., for 12, the pairs are 1 and 12, 2 and 6, 3 and 4). Perfect squares like 9 only have the divisor 3 once, making the count odd.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Brainteaser.

Companies

Asked at: Accenture.

Bulb Switcher | LeetCode 319

Medium

Math Brainteaser

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

Solution Code

Available in:

class Solution {
    public int bulbSwitch(int n) {
        return (int)Math.sqrt(n);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 319: Bulb Switcher - Detailed explanation and solution approaches.

Topics

Math Brainteaser

Companies

Accenture

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:319

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

2 steps

1
Step 1: Calculate the integer square root of 'n'. This effectively finds the largest integer 'x' such that x*x <= n.
2
Step 2: Return the integer square root. This value represents the count of perfect squares from 1 to 'n' (1*1, 2*2, 3*3, ... x*x).

💡

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 uses a constant number of operations (calculating the square root) regardless of the input 'n'. Therefore, the time complexity is O(1). It also uses a constant amount of extra space, 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

Iterative Toggling

Simulate the toggling process round by round using an array to represent the bulbs. This directly implements the problem description.

Time: O(n^2)Space: O(n)

Trade-offs: This approach is much slower for larger values of 'n' because it performs n toggling rounds. However, it's more intuitive and easier to understand for someone unfamiliar with the mathematical insight.

✨

Key Insights

1Insight 1: A bulb will be ON at the end if it has been toggled an odd number of times. It will be OFF if toggled an even number of times.
2Insight 2: The number of times a bulb is toggled is equal to the number of its divisors. For example, bulb number 12 is toggled in rounds 1, 2, 3, 4, 6, and 12 (6 divisors).
3Insight 3: A number has an odd number of divisors if and only if it is a perfect square. For example, 9 has divisors 1, 3, and 9 (3 divisors). If a number is not a perfect square, its divisors come in pairs (e.g., for 12, the pairs are 1 and 12, 2 and 6, 3 and 4). Perfect squares like 9 only have the divisor 3 once, making the count odd.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n = 0. The integer square root of 0 is 0, which is the correct answer because there are no bulbs to toggle, and therefore none are on.
Edge case 2: n = 1. The integer square root of 1 is 1, which means only the first bulb is on after 1 round, and this is the correct answer.
Edge case 3: Large values of 'n'. The `sqrt` function in the provided code will handle large inputs without causing overflow issues because it's a built-in function designed for such scenarios.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Directly implementing the toggling process without recognizing the pattern. This leads to an O(n^2) solution, which will time out for large input values.

✗Common mistake 2: Failing to consider that a number has an odd number of divisors if and only if it's a perfect square. Not realizing this mathematical fact can prevent finding the efficient O(1) solution.

✅

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 is a good example of how mathematical insights can dramatically simplify algorithms and improve their efficiency. It highlights the importance of recognizing patterns.

🏦Real-world scenario 2: Though the problem itself is abstract, the underlying principle of finding the number of perfect squares within a given range has applications in number theory and cryptography.

💡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: October 18, 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 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

Frequently Asked Questions

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 "Bulb Switcher"?

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

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