Minimum Bit Flips to Convert Number : When working with binary numbers in programming, you might encounter a problem where you need to find the minimum number of bit flips required to convert one number into another. This is a common interview and LeetCode problem that tests your understanding of bit manipulation.

In this guide, we’ll explain the concept step-by-step, walk through examples, and provide a LeetCode-style C++ solution.

What Does “Bit Flip” Mean?

A bit flip means changing a 0 to 1 or a 1 to 0 in a number’s binary representation.

Example:

• Original bit: 0 → Flipped bit: 1
• Original bit: 1 → Flipped bit: 0

If you have two integers, finding the minimum bit flips means determining the positions where the bits differ.

Understanding the Problem

Problem Statement:
Given two integers start and goal, return the minimum number of bit flips required to convert start to goal.

Example

Input: start = 10, goal = 7
Binary of 10 → 1010
Binary of 7  → 0111

Comparing bit by bit:

1 0 1 0  
0 1 1 1  
↑ ↑   ↑  
Different bits at positions 1, 2, and 4 → 3 flips needed

Approach

1. Use XOR to identify different bits:
   • XOR (^) returns 1 for differing bits, 0 for same bits.
2. Count set bits in the XOR result:
   • The number of 1s = the number of bit flips needed.

Step-by-Step Example

For start = 29 and goal = 15:

1. XOR Operation: 29 (11101) 15 (01111) XOR → 10010
2. Count set bits in 10010:
   There are 2 set bits → 2 flips needed.

1. C++ Solution

#include <iostream>
using namespace std;

int minBitFlips(int start, int goal) {
    int xorValue = start ^ goal; // Step 1: XOR to find differing bits
    int count = 0;
    while (xorValue > 0) {
        count += xorValue & 1;  // Step 2: Count set bits
        xorValue >>= 1;
    }
    return count;
}

int main() {
    int start = 10, goal = 7;
    cout << "Minimum Bit Flips: " << minBitFlips(start, goal) << endl;
    return 0;
}

✅ Shortcut in C++ (GCC/Clang):

return __builtin_popcount(start ^ goal);

2. C Solution for Minimum Bit Flips to Convert Number

#include <stdio.h>

int minBitFlips(int start, int goal) {
    int xorValue = start ^ goal; // XOR to find differing bits
    int count = 0;
    while (xorValue > 0) {
        count += xorValue & 1;  // Count set bits
        xorValue >>= 1;
    }
    return count;
}

int main() {
    int start = 10, goal = 7;
    printf("Minimum Bit Flips: %d\n", minBitFlips(start, goal));
    return 0;
}

3. Java Solution for Minimum Bit Flips to Convert Number

public class Main {
    public static int minBitFlips(int start, int goal) {
        int xorValue = start ^ goal; // XOR to find differing bits
        int count = 0;
        while (xorValue > 0) {
            count += xorValue & 1; // Count set bits
            xorValue >>= 1;
        }
        return count;
    }

    public static void main(String[] args) {
        int start = 10, goal = 7;
        System.out.println("Minimum Bit Flips: " + minBitFlips(start, goal));
    }
}

✅ Shortcut in Java (Java 8+):

return Integer.bitCount(start ^ goal);

4. Python Solution for Minimum Bit Flips to Convert Number

def min_bit_flips(start, goal):
    xor_value = start ^ goal  # XOR to find differing bits
    count = 0
    while xor_value > 0:
        count += xor_value & 1  # Count set bits
        xor_value >>= 1
    return count

# Example
start = 10
goal = 7
print("Minimum Bit Flips:", min_bit_flips(start, goal))

✅ Shortcut in Python:

return bin(start ^ goal).count('1')

Time Complexity for All Solutions

• O(k), where k = number of bits in the integer.

Real-World Applications

• Error Detection & Correction in data transmission.
• Cryptography and data encoding.
• Digital Circuit Design optimization.

Key Takeaways

• Use XOR to detect differences in bits.
• Count the number of 1s in the XOR result to get the flips.
• This is a popular LeetCode problem for practicing bit manipulation.

FAQ for Minimum Bit Flips to Convert Number

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