What is the time complexity of Convert to Base -2?

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

What is the space complexity of Convert to Base -2?

The space complexity is O(logN).

What data structures are used to solve Convert to Base -2?

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

What are the key insights for solving Convert to Base -2?

Insight 1: The core idea is to repeatedly divide the number by -2 and keep track of the remainders. Since remainders can only be 0 or 1, the problem effectively converts to finding these binary digits. Insight 2: The division by -2 can result in negative quotients. We can leverage the bitwise AND operator (`& 1`) to extract the least significant bit (LSB), which will be the remainder (0 or 1).

Which companies ask LeetCode 1017. Convert to Base -2 in interviews?

This problem is frequently asked at Airbnb.

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Convert to Base -2 - LeetCode 1017 Solution

Q: How to solve Convert to Base -2?

The solution uses an iterative approach. It repeatedly divides the given number `n` by -2. The remainder of each division (which will be either 0 or 1) becomes a digit in the base -2 representation. The digits are collected in reverse order (least significant digit first), and then reversed at the end to form the final result. The bitwise AND operator `& 1` is used to efficiently extract the remainder when dividing by -2. The loop continues until the number becomes 0. A special case is handled when the input `n` is 0, where the output is directly "0".

Convert to Base -2 - Complete Solution Guide

Convert to Base -2 is LeetCode problem 1017, 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

Given an integer n , return a binary string representing its representation in base -2 . Note that the returned string should not have leading zeros unless the string is "0" . Example 1: Input: n = 2 Output: "110" Explantion: (-2) 2 + (-2) 1 = 2 Example 2: Input: n = 3 Output: "111" Explantion: (-2) 2 + (-2) 1 + (-2) 0 = 3 Example 3: Input: n = 4 Output: "100" Explantion: (-2) 2 = 4 Constraints: 0 <= n <= 10 9

Detailed Explanation

The problem asks us to convert a given integer `n` into its equivalent representation in base -2. The output should be a string representing this base -2 number. Leading zeros should be avoided, except when the input is 0, in which case the output should be "0". The input `n` is constrained to be between 0 and 10^9 inclusive.

Solution Approach

The solution uses an iterative approach. It repeatedly divides the given number `n` by -2. The remainder of each division (which will be either 0 or 1) becomes a digit in the base -2 representation. The digits are collected in reverse order (least significant digit first), and then reversed at the end to form the final result. The bitwise AND operator `& 1` is used to efficiently extract the remainder when dividing by -2. The loop continues until the number becomes 0. A special case is handled when the input `n` is 0, where the output is directly "0".

Step-by-Step Algorithm

Step 1: Handle the base case where n is 0. If n is 0, return "0".
Step 2: Initialize an empty string or list to store the base -2 digits.
Step 3: While n is not equal to 0, repeat the following steps:
Step 4: Extract the least significant bit (LSB) of n using the bitwise AND operator (`n & 1`). Convert this bit to a string and append it to the result string/list.
Step 5: Divide n by -2. This is done using `n = -(n >> 1)`. The right shift operator (`>> 1`) effectively divides by 2, and the negation handles the base -2 aspect.
Step 6: After the loop finishes, reverse the string/list to obtain the correct order of digits.
Step 7: Return the reversed string.

Key Insights

Insight 1: The core idea is to repeatedly divide the number by -2 and keep track of the remainders. Since remainders can only be 0 or 1, the problem effectively converts to finding these binary digits.
Insight 2: The division by -2 can result in negative quotients. We can leverage the bitwise AND operator (`& 1`) to extract the least significant bit (LSB), which will be the remainder (0 or 1).
Insight 3: The negative division needs careful handling. We use `n = -(n >> 1)` to propagate the negative sign properly, which is crucial for obtaining the correct representation.

Complexity Analysis

Time Complexity: O(logN)

Space Complexity: O(logN)

Topics

This problem involves: Math.

Companies

Asked at: Airbnb.

Convert to Base -2 | LeetCode 1017

Medium

Math

Time: O(logN)Space: O(logN)

Solution Code

Available in:

class Solution {
    public String baseNeg2(int n) {
        if (n == 0) {
            return "0";
        }
        
        StringBuilder res = new StringBuilder();
        while (n != 0) {
            res.append(n & 1);
            n = -(n >> 1);
        }
        
        return res.reverse().toString();
    }
}

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Problem Info

Description

Problem 1017: Convert to Base -2 - Detailed explanation and solution approaches.

Topics

Math

Companies

Airbnb

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1017

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: Handle the base case where n is 0. If n is 0, return "0".
2
Step 2: Initialize an empty string or list to store the base -2 digits.
3
Step 3: While n is not equal to 0, repeat the following steps:
4
Step 4: Extract the least significant bit (LSB) of n using the bitwise AND operator (`n & 1`). Convert this bit to a string and append it to the result string/list.
5
Step 5: Divide n by -2. This is done using `n = -(n >> 1)`. The right shift operator (`>> 1`) effectively divides by 2, and the negation handles the base -2 aspect.
6
Step 6: After the loop finishes, reverse the string/list to obtain the correct order of digits.
7
Step 7: Return the reversed string.

💡

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(logN)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(logN)

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The number of iterations of the while loop is proportional to the number of digits in the base -2 representation of n. The number of digits needed to represent n in base -2 is logarithmic with respect to n. Therefore, the time complexity is O(logN). The space complexity is also O(logN) because we store the base -2 digits in a string/list, and the length of this string/list is proportional to the number of digits, which is logarithmic with respect to n.

📖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

Recursive Approach

A recursive version can be implemented, mirroring the iterative approach. The base case would be n=0, and the recursive step would involve calculating the LSB and recursively calling the function with the quotient.

Time: O(logN)Space: O(logN) due to the call stack

Trade-offs: The recursive approach might be slightly less efficient due to function call overhead. However, it can be more concise and readable for some programmers.

✨

Key Insights

1Insight 1: The core idea is to repeatedly divide the number by -2 and keep track of the remainders. Since remainders can only be 0 or 1, the problem effectively converts to finding these binary digits.
2Insight 2: The division by -2 can result in negative quotients. We can leverage the bitwise AND operator (`& 1`) to extract the least significant bit (LSB), which will be the remainder (0 or 1).
3Insight 3: The negative division needs careful handling. We use `n = -(n >> 1)` to propagate the negative sign properly, which is crucial for obtaining the correct representation.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n = 0. The code explicitly handles this case by returning "0". This is crucial because the general algorithm wouldn't terminate correctly for n = 0.
Edge case 2: n = 1. The algorithm correctly computes the result as "1".
Edge case 3: Large values of n (close to 10^9). The algorithm is designed to handle these values correctly due to the properties of bitwise operations and integer division.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the negative division. The `-(n >> 1)` operation is critical. Simply using `n >> 1` (or equivalent integer division) will not give the correct base -2 representation.

✗Common mistake 2: Not reversing the string. The digits are generated in reverse order, so reversing the string is necessary to obtain the correct representation.

✗Common mistake 3: Not handling the n=0 edge case. Without a specific check for n=0, the while loop will not execute, leading to an incorrect result (likely an empty string).

✗Common mistake 4: Integer overflow. In languages like C/C++, using too small an allocated memory could cause a buffer overflow, which is unsafe. Ensure sufficient space is allocated beforehand.

✅

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: Computer architecture. Non-standard bases like base -2 (or other negative bases) can theoretically be used in specialized computer architectures. While not widely used, they offer unique mathematical properties.

🏦Real-world scenario 2: Cryptography. Certain cryptographic algorithms leverage mathematical properties of different number systems. Although base -2 is not directly used, the underlying principles of number representation and conversion are fundamental concepts.

💡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

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

⚠️ 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 26, 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 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

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 "Convert to Base -2"?

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

🤖 Need help with this problem?