What is the time complexity of Minimum Bit Flips to Convert Number?

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

What data structures are used to solve Minimum Bit Flips to Convert Number?

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

What are the key insights for solving Minimum Bit Flips to Convert Number?

Insight 1: The XOR operation (`^`) is crucial. `start ^ goal` gives a number where each bit is 1 if the corresponding bits in `start` and `goal` are different, and 0 otherwise. This directly represents the bits that need flipping. Insight 2: Counting set bits (bits with value 1) in the XOR result efficiently determines the minimum number of bit flips. Iterating through the bits and checking each one is inefficient. We can use bitwise operations for efficiency.

Which companies ask LeetCode 2220. Minimum Bit Flips to Convert Number in interviews?

This problem is frequently asked at persistent systems.

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Minimum Bit Flips to Convert Number - LeetCode 2220 Solution

Minimum Bit Flips to Convert Number - Complete Solution Guide

Minimum Bit Flips to Convert Number is LeetCode problem 2220, 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

A bit flip of a number x is choosing a bit in the binary representation of x and flipping it from either 0 to 1 or 1 to 0 . For example, for x = 7 , the binary representation is 111 and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get 110 , flip the second bit from the right to get 101 , flip the fifth bit from the right (a leading zero) to get 10111 , etc. Given two integers start and goal , return the minimum number of b

Detailed Explanation

The problem asks to find the minimum number of bit flips required to convert one integer ( `start` ) into another ( `goal` ). A bit flip involves changing a single bit from 0 to 1 or vice versa in the binary representation of a number. The solution should handle any non-negative integers as input.

Solution Approach

The provided solution uses a bitwise approach. First, it calculates the XOR of `start` and `goal`. This XOR result contains 1s in the bit positions where `start` and `goal` differ. Then, it iteratively counts the number of 1s in the XOR result using bit manipulation. The loop continues until all 1s are counted (the XOR result becomes 0). The count represents the minimum number of bit flips needed.

Step-by-Step Algorithm

Step 1: Calculate the XOR of `start` and `goal` using the bitwise XOR operator (`^`). Store the result in a variable (e.g., `xor`).
Step 2: Initialize a counter variable (`count`) to 0. This will store the number of bit flips.
Step 3: While `xor` is greater than 0:
Step 4: Check the least significant bit (LSB) of `xor` using the bitwise AND operator (`&`). If the LSB is 1, increment `count`.
Step 5: Right-shift `xor` by 1 bit using the right-shift operator (`>>`) to move to the next bit.
Step 6: Repeat steps 3-5 until `xor` becomes 0.
Step 7: Return the value of `count` which represents the minimum number of bit flips.

Key Insights

Insight 1: The XOR operation (`^`) is crucial. `start ^ goal` gives a number where each bit is 1 if the corresponding bits in `start` and `goal` are different, and 0 otherwise. This directly represents the bits that need flipping.
Insight 2: Counting set bits (bits with value 1) in the XOR result efficiently determines the minimum number of bit flips. Iterating through the bits and checking each one is inefficient. We can use bitwise operations for efficiency.
Insight 3: The problem is effectively calculating the Hamming distance between two integers. The Hamming distance is the number of positions at which the corresponding bits are different.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Bit Manipulation.

Companies

Asked at: persistent systems.

Minimum Bit Flips to Convert Number | LeetCode 2220

Easy

Bit Manipulation

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

Solution Code

Available in:

class Solution {
    public int minBitFlips(int start, int goal) {
        int xor = start ^ goal;
        int count = 0;
        while (xor > 0) {
            count += xor & 1;
            xor >>= 1;
        }
        return count;
    }
}

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Problem Info

Description

Problem 2220: Minimum Bit Flips to Convert Number - Detailed explanation and solution approaches.

Topics

Bit Manipulation

Companies

persistent systems

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2220

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: Calculate the XOR of `start` and `goal` using the bitwise XOR operator (`^`). Store the result in a variable (e.g., `xor`).
2
Step 2: Initialize a counter variable (`count`) to 0. This will store the number of bit flips.
3
Step 3: While `xor` is greater than 0:
4
Step 4: Check the least significant bit (LSB) of `xor` using the bitwise AND operator (`&`). If the LSB is 1, increment `count`.
5
Step 5: Right-shift `xor` by 1 bit using the right-shift operator (`>>`) to move to the next bit.
6
Step 6: Repeat steps 3-5 until `xor` becomes 0.
7
Step 7: Return the value of `count` which represents the minimum number of bit flips.

💡

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 time complexity is O(log n) because the while loop iterates at most log₂(n) times, where n is the maximum of `start` and `goal`. This is because the number of bits in the binary representation of an integer is proportional to its logarithm base 2. The space complexity is O(1) because only a constant number of variables are used 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

Built-in functions (count set bits)

Some languages (like Python) have built-in functions to count set bits (e.g., `bin(num).count('1')`). This can slightly simplify the code but might not be as efficient as bitwise operations in some cases.

Time: O(log n) or O(b) where b is the number of bits, which is technically O(log n)Space: O(1) or O(log n) depending on the implementation of the built-in function

Trade-offs: Using built-in functions can improve code readability but may slightly reduce performance compared to direct bit manipulation. The choice depends on readability vs performance priorities.

✨

Key Insights

1Insight 1: The XOR operation (`^`) is crucial. `start ^ goal` gives a number where each bit is 1 if the corresponding bits in `start` and `goal` are different, and 0 otherwise. This directly represents the bits that need flipping.
2Insight 2: Counting set bits (bits with value 1) in the XOR result efficiently determines the minimum number of bit flips. Iterating through the bits and checking each one is inefficient. We can use bitwise operations for efficiency.
3Insight 3: The problem is effectively calculating the Hamming distance between two integers. The Hamming distance is the number of positions at which the corresponding bits are different.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `start` and `goal` are equal. In this case, the XOR will be 0, and the function will correctly return 0.
Edge case 2: `start` or `goal` is 0. The solution correctly handles this situation, counting the bits that need to be flipped to reach the other number.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using the XOR operator efficiently. Beginners might try to convert the numbers to binary strings, compare them character by character, and count the differences. This is less efficient.

✗Common mistake 2: Incorrectly handling the bitwise operations. Errors can occur if the AND and right-shift operators are not used correctly. For example, using a left shift or an incorrect bitmask.

✅

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: Error detection and correction in data transmission. Hamming distance is fundamental to error-correcting codes which detect and correct bit errors in data transmitted over noisy channels.

🏦Real-world scenario 2: DNA sequencing and comparison. Comparing DNA sequences can involve calculating the Hamming distance to assess the similarity between two sequences.

💡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

XOR is your friend for "find the single number" problems. Remember: n & (n-1) clears the lowest set bit.

🎯 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.

👨‍💻 From Experience

This is one of the first problems I solved when learning Bit Manipulation. It's a great foundation for more complex problems.

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: July 31, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 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 "Minimum Bit Flips to Convert Number"?

For this easy-level Bit Manipulation 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.

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