What is the time complexity of Power of Four?

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

What data structures are used to solve Power of Four?

This problem uses Math, Bit Manipulation, Recursion. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Power of Four?

Insight 1: Powers of four are also powers of two. This allows us to use bit manipulation techniques more efficiently. Only one bit will be set in the binary representation of a power of four. Insight 2: Bit manipulation techniques are significantly faster than iterative or recursive approaches, allowing for constant time complexity solutions. Analyzing the binary representation is crucial.

Which companies ask LeetCode 342. Power of Four in interviews?

This problem is frequently asked at Qualcomm, Two Sigma, Wix.

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Power of Four - LeetCode 342 Solution

Q: How to solve Power of Four?

The provided solutions employ different strategies. The Python and Java solutions use an iterative approach to repeatedly divide by 4 until either the result is 1 (indicating a power of four) or the remainder is non-zero (indicating it's not a power of four). The C++ solution cleverly uses bit manipulation: it checks if only one bit is set (meaning it's a power of two) and then verifies that it's a power of four by checking the pattern of the set bit's position. The C solution also uses bit manipulation, specifically checking if the number is a power of 2 and then examining the specific bit pattern characteristic of powers of four.

Power of Four - Complete Solution Guide

Power of Four is LeetCode problem 342, 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

Given an integer n , return true if it is a power of four. Otherwise, return false . An integer n is a power of four, if there exists an integer x such that n == 4 x . Example 1: Input: n = 16 Output: true Example 2: Input: n = 5 Output: false Example 3: Input: n = 1 Output: true Constraints: -2 31 <= n <= 2 31 - 1 Follow up: Could you solve it without loops/recursion?

Detailed Explanation

The problem asks to determine if a given integer `n` is a power of four. This means checking if there exists an integer `x` such that `n == 4^x`. The input is an integer `n` within the range of a 32-bit signed integer. The output is a boolean value: `true` if `n` is a power of four, and `false` otherwise. The problem specifically encourages finding a solution without using loops or recursion.

Solution Approach

The provided solutions employ different strategies. The Python and Java solutions use an iterative approach to repeatedly divide by 4 until either the result is 1 (indicating a power of four) or the remainder is non-zero (indicating it's not a power of four). The C++ solution cleverly uses bit manipulation: it checks if only one bit is set (meaning it's a power of two) and then verifies that it's a power of four by checking the pattern of the set bit's position. The C solution also uses bit manipulation, specifically checking if the number is a power of 2 and then examining the specific bit pattern characteristic of powers of four.

Step-by-Step Algorithm

Step 1: Handle edge cases. Check if `n` is less than or equal to 0. If so, return `false` because powers of four are always positive.
Step 2 (Iterative approach): Repeatedly divide `n` by 4 using integer division (`//` in Python, `/` in Java). If the remainder is ever non-zero, `n` is not a power of four, so return `false`. If after repeated division, `n` becomes 1, then it is a power of four, so return `true`.
Step 2 (Bit manipulation approach): Check if `n` is a power of two using `(n & (n - 1)) == 0`. If not, return `false`. Then, check the specific bit pattern characteristic of powers of four (e.g., only one bit is set and its position has specific properties).
Step 3: Return the result (true or false) based on the checks.

Key Insights

Insight 1: Powers of four are also powers of two. This allows us to use bit manipulation techniques more efficiently. Only one bit will be set in the binary representation of a power of four.
Insight 2: Bit manipulation techniques are significantly faster than iterative or recursive approaches, allowing for constant time complexity solutions. Analyzing the binary representation is crucial.
Insight 3: Handling edge cases like negative numbers and zero is important. Powers of four are always positive.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Math, Bit Manipulation, Recursion.

Companies

Asked at: Qualcomm, Two Sigma, Wix.

Power of Four | LeetCode 342

Easy

Math Bit Manipulation Recursion

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

Solution Code

Available in:

class Solution {
    public boolean isPowerOfFour(int n) {
        if(n <= 0) return false;
        while(n % 4 == 0){
            n /= 4;
        }
        return n == 1;
    }
}

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Problem Info

Description

Problem 342: Power of Four - Detailed explanation and solution approaches.

Topics

Math Bit Manipulation Recursion

Companies

QualcommTwo SigmaWix

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:342

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

4 steps

1
Step 1: Handle edge cases. Check if `n` is less than or equal to 0. If so, return `false` because powers of four are always positive.
2
Step 2 (Iterative approach): Repeatedly divide `n` by 4 using integer division (`//` in Python, `/` in Java). If the remainder is ever non-zero, `n` is not a power of four, so return `false`. If after repeated division, `n` becomes 1, then it is a power of four, so return `true`.
3
Step 2 (Bit manipulation approach): Check if `n` is a power of two using `(n & (n - 1)) == 0`. If not, return `false`. Then, check the specific bit pattern characteristic of powers of four (e.g., only one bit is set and its position has specific properties).
4
Step 3: Return the result (true or false) based on the checks.

💡

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 iterative approach has a logarithmic time complexity because the number of divisions is proportional to the logarithm base 4 of n. The bit manipulation approach has constant time complexity because bitwise operations take constant time regardless of the input size.

📖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

Using logarithms

Calculate the base-4 logarithm of `n`. If the result is an integer, then `n` is a power of four.

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

Trade-offs: This approach might be less efficient than bit manipulation due to the overhead of logarithm calculations. It's also less readable.

✨

Key Insights

1Insight 1: Powers of four are also powers of two. This allows us to use bit manipulation techniques more efficiently. Only one bit will be set in the binary representation of a power of four.
2Insight 2: Bit manipulation techniques are significantly faster than iterative or recursive approaches, allowing for constant time complexity solutions. Analyzing the binary representation is crucial.
3Insight 3: Handling edge cases like negative numbers and zero is important. Powers of four are always positive.

📚

Key Terms Explained

For Students

Recursion

A function that calls itself to solve smaller versions of the same problem.

Edge Cases

Edge case 1: n = 0. The code correctly handles this by returning `false`.
Edge case 2: n = 1. This is a power of four (4^0), and the code correctly returns `true`.
Edge case 3: n = Negative Numbers. The code correctly handles negative numbers by returning `false`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case of `n` being 0 or negative. Powers of four are always positive.

✗Common mistake 2: Incorrectly checking for powers of two. A power of four must also be a power of two, but not all powers of two are powers of four. Efficiently identifying this is key.

✅

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: Checking for valid data structures sizes in systems where memory allocation is restricted to powers of four.

🏦Real-world scenario 2: In bit manipulation algorithms for image processing or compression, identifying specific patterns in data representations which rely on powers of four.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: November 10, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Power of Four"?

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

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