What is the time complexity of Number of Bit Changes to Make Two Integers Equal?

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

What is the space complexity of Number of Bit Changes to Make Two Integers Equal?

The space complexity is O(log(n)).

What data structures are used to solve Number of Bit Changes to Make Two Integers Equal?

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

What are the key insights for solving Number of Bit Changes to Make Two Integers Equal?

Insight 1: The problem can be efficiently solved using bit manipulation. Comparing the bits of `n` and `k` directly allows for a straightforward approach. Insight 2: The XOR operation (`^`) is crucial. `n ^ k` gives a number where each bit is 1 if the corresponding bits in `n` and `k` are different and 0 if they are the same. This efficiently identifies the bits that need changing.

Which companies ask LeetCode 3226. Number of Bit Changes to Make Two Integers Equal in interviews?

This problem is frequently asked at ThoughtWorks.

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Number of Bit Changes to Make Two Integers Equal - LeetCode 3226 Solution

Number of Bit Changes to Make Two Integers Equal - Complete Solution Guide

Number of Bit Changes to Make Two Integers Equal is LeetCode problem 3226, 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

You are given two positive integers n and k . You can choose any bit in the binary representation of n that is equal to 1 and change it to 0. Return the number of changes needed to make n equal to k . If it is impossible, return -1. Example 1: Input: n = 13, k = 4 Output: 2 Explanation: Initially, the binary representations of n and k are n = (1101) 2 and k = (0100) 2 . We can change the first and fourth bits of n . The resulting integer is n = ( 0 10 0 ) 2 = k . Example 2: Input: n = 21, k = 21

Detailed Explanation

The problem asks you to find the minimum number of bit changes required to transform a positive integer `n` into another positive integer `k`. The only allowed operation is changing a '1' bit in `n` to a '0' bit. If it's impossible to make `n` equal to `k` using this operation, the function should return -1. The input consists of two positive integers, `n` and `k`, and the output is a non-negative integer representing the number of changes or -1 if it's impossible.

Solution Approach

The provided solutions utilize bit manipulation to efficiently determine the minimum number of changes. They iterate through the bits of `n` and `k` (or use bitwise operations on `n ^ k`) and count the number of times a '1' bit in `n` needs to be changed to a '0' bit to match `k`. If a '0' in `n` needs to become a '1' in `k`, it's impossible, and -1 is returned. The Python solution converts to binary strings for clarity, while the Java, C++, and C solutions use bitwise operations for efficiency.

Step-by-Step Algorithm

Step 1: Calculate `n ^ k` (XOR of n and k). This highlights bits that differ between n and k.
Step 2: Iterate through the bits of `n ^ k` (or the bits of n and k simultaneously).
Step 3: For each bit, if it's 1 in `n ^ k`, check if the corresponding bit in `n` is 1. If so, increment the change counter. If the corresponding bit in `n` is 0, it's impossible to achieve `k` from `n`, so return -1.
Step 4: If the loop completes without returning -1, return the change counter.

Key Insights

Insight 1: The problem can be efficiently solved using bit manipulation. Comparing the bits of `n` and `k` directly allows for a straightforward approach.
Insight 2: The XOR operation (`^`) is crucial. `n ^ k` gives a number where each bit is 1 if the corresponding bits in `n` and `k` are different and 0 if they are the same. This efficiently identifies the bits that need changing.
Insight 3: It's impossible to transform `n` into `k` if any bit is 1 in `k` but 0 in `n`. This condition must be checked to correctly handle cases where transformation is not possible.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(log(n))

Topics

This problem involves: Bit Manipulation.

Companies

Asked at: ThoughtWorks.

Number of Bit Changes to Make Two Integers Equal | LeetCode 3226

Easy

Bit Manipulation

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

Solution Code

Available in:

class Solution {
    public int minChanges(int n, int k) {
        int xor = n ^ k;
        int count = 0;
        if (n < k) {
            return -1;
        }
        while (xor > 0) {
            if ((xor & 1) == 1) {
                if ((n & 1) == 1) {
                    count++;
                } else {
                    return -1;
                }
            }
            n >>= 1;
            k >>= 1;
            xor >>= 1;
        }
        return count;
    }
}

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Problem Info

Description

Problem 3226: Number of Bit Changes to Make Two Integers Equal - Detailed explanation and solution approaches.

Topics

Bit Manipulation

Companies

ThoughtWorks

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3226

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: Calculate `n ^ k` (XOR of n and k). This highlights bits that differ between n and k.
2
Step 2: Iterate through the bits of `n ^ k` (or the bits of n and k simultaneously).
3
Step 3: For each bit, if it's 1 in `n ^ k`, check if the corresponding bit in `n` is 1. If so, increment the change counter. If the corresponding bit in `n` is 0, it's impossible to achieve `k` from `n`, so return -1.
4
Step 4: If the loop completes without returning -1, return the change counter.

💡

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(log(n))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is O(log n) because the number of bits in the binary representation of n (and k) is proportional to log₂(n). The space complexity is O(1) because we only use a constant amount of extra space to store variables like `changes` and `xor_val`. The iterative approach directly processes the bits without using additional data structures that scale with 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

String Conversion and Comparison

Convert both `n` and `k` to their binary string representations. Then, iterate through the strings and compare bits. Count changes while checking for impossible conditions. This method is less efficient than bitwise operations.

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

Trade-offs: This approach is less efficient than the bitwise operations but might be easier to understand for beginners. It uses more space due to creating string representations. The bitwise approach is preferred for its efficiency and conciseness.

✨

Key Insights

1Insight 1: The problem can be efficiently solved using bit manipulation. Comparing the bits of `n` and `k` directly allows for a straightforward approach.
2Insight 2: The XOR operation (`^`) is crucial. `n ^ k` gives a number where each bit is 1 if the corresponding bits in `n` and `k` are different and 0 if they are the same. This efficiently identifies the bits that need changing.
3Insight 3: It's impossible to transform `n` into `k` if any bit is 1 in `k` but 0 in `n`. This condition must be checked to correctly handle cases where transformation is not possible.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n < k. This case is explicitly handled. It's impossible to transform a smaller number into a larger number by only changing 1s to 0s, so -1 is returned.
Edge case 2: n == k. In this case, no changes are needed, and the function correctly returns 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if a '0' bit in n needs to become a '1' in k. This leads to incorrect results when transformation is impossible.

✗Common mistake 2: Inefficiently converting numbers to binary using loops instead of leveraging bitwise operators. Bitwise operations provide significantly better performance.

✅

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: Digital signal processing. When dealing with binary data representing signals, you might need to determine the minimal changes to correct errors or align signals.

🏦Real-world scenario 2: Genetic algorithms. Representing genes as binary strings, this algorithm could help to measure the 'distance' or number of changes between different gene 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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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 14, 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 13

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

@InterviewProJul 6

❓

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 "Number of Bit Changes to Make Two Integers Equal"?

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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