Count number of bits to be flipped to convert A to B

Given two numbers A and B. Write a program to count the number of bits needed to be flipped to convert A to B. 

Examples: 

Input: A = 10, B = 20
Output: 4
Explanation: Binary representation of A is 00001010
Binary representation of B is 00010100
We need to flip highlighted four bits in A to make it B.

Input: A = 7, B = 10
Output: 3
Explanation: Binary representation of A is 00000111
Binary representation of B is 00001010
We need to flip highlighted three bits in A to make it B.

Count the number of bits to be flipped to convert A to B using the XOR operator:

To solve the problem follow the below idea:

Calculate (A XOR B), since 0 XOR 1 and 1 XOR 0 is equal to 1. So calculating the number of set bits in A XOR B will give us the count of the number of unmatching bits in A and B, which needs to be flipped

Follow the given steps to solve the problem:

• Calculate the XOR of A and B
• Count the set bits in the above-calculated XOR result
• Return the count

Below is the implementation of the above approach:

C

// C program for the above approach
#include <stdio.h>

// Function that count set bits
int countSetBits(int n)
{
    int count = 0;
    while (n > 0) {
        count++;
        n &= (n - 1);
    }
    return count;
}

// Function that return count of flipped number
int FlippedCount(int a, int b)
{
    // Return count of set bits in a XOR b
    return countSetBits(a ^ b);
}

// Driver code
int main()
{
    int a = 10;
    int b = 20;

      // Function call
    printf("%d\n", FlippedCount(a, b));
    return 0;
}

// This code is contributed by Sania Kumari Gupta
// (kriSania804)

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that count set bits
int countSetBits(int n)
{
    int count = 0;
    while (n > 0) {
        count++;
        n &= (n - 1);
    }
    return count;
}

// Function that return count of
// flipped number
int FlippedCount(int a, int b)
{
    // Return count of set bits in
    // a XOR b
    return countSetBits(a ^ b);
}

// Driver code
int main()
{
    int a = 10;
    int b = 20;

      // Function call
    cout << FlippedCount(a, b) << endl;
    return 0;
}

// This code is contributed by Sania Kumari Gupta
// (kriSania804)

Java

// Java program for the above approach
import java.util.*;

class Count {

    // Function that count set bits
    public static int countSetBits(int n)
    {
        int count = 0;
        while (n != 0) {
            count++;
            n &= (n - 1);
        }
        return count;
    }

    // Function that return count of
    // flipped number
    public static int FlippedCount(int a, int b)
    {
        // Return count of set bits in
        // a XOR b
        return countSetBits(a ^ b);
    }

    // Driver code
    public static void main(String[] args)
    {
        int a = 10;
        int b = 20;

          // Function call
        System.out.print(FlippedCount(a, b));
    }
}

// This code is contributed by rishabh_jain

Python3

# Python3 program for the above approach

# Function that count set bits


def countSetBits(n):
    count = 0
    while n:
        count += 1
        n &= (n-1)
    return count

# Function that return count of
# flipped number


def FlippedCount(a, b):

    # Return count of set bits in
    # a XOR b
    return countSetBits(a ^ b)


# Driver code
if __name__ == "__main__":
  a = 10
  b = 20

  # Function call
  print(FlippedCount(a, b))

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program for the above approach

using System;

class Count {

    // Function that count set bits
    public static int countSetBits(int n)
    {
        int count = 0;
        while (n != 0) {
            count++;
            n &= (n - 1);
        }
        return count;
    }

    // Function that return
    // count of flipped number
    public static int FlippedCount(int a, int b)
    {
        // Return count of set
        // bits in a XOR b
        return countSetBits(a ^ b);
    }

    // Driver code
    public static void Main()
    {
        int a = 10;
        int b = 20;

          // Function call
        Console.WriteLine(FlippedCount(a, b));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// php program for the above approach

// Function that count set bits
function countSetBits($n)
{
    $count = 0;
    while($n)
    {
        $count += 1;
        $n &= (n-1);
    }
    return $count;
}

// Function that return
// count of flipped number
function FlippedCount($a, $b)
{
    // Return count of set
    // bits in a XOR b
    return countSetBits($a ^ $b);
}

// Driver code
$a = 10;
$b = 20;

// Function call
echo FlippedCount($a, $b);

// This code is contributed by mits
?>

Javascript

// Count number of bits to be flipped
// to convert A into Bclass Count {

    // Function that count set bits
    function countSetBits(n) {
        var count = 0;
        while (n != 0) {
            count++;
            n &= (n - 1);
        }
        return count;
    }

    // Function that return count of
    // flipped number
    function FlippedCount(a , b) {
        // Return count of set bits in
        // a XOR b
        return countSetBits(a ^ b);
    }

    // Driver code
        var a = 10;
        var b = 20;
        document.write(FlippedCount(a, b));

// This code is contributed by shikhasingrajput

Output

4

Time Complexity: O(K) where K is the number of bits
Auxiliary Space: O(1)

Note: Set bits in (a XOR b) can also be computer using built in function __builtin_popcount() in C/C++

Below is the implementation of the above approach:

C++

// C++ program to Count number of bits to be flipped
// to convert A into B
#include <iostream>
using namespace std;

// Driver code
int main()
{
    int a = 10;
    int b = 20;

      // Function call
    cout <<__builtin_popcount(a^b) << endl;
    return 0;
}

// This code is contributed by Suruchi Kumari

C

//C program to Count number of bits to be flipped
// to convert A into B
#include <stdio.h>

// Driver code
int main()
{
    int a = 10;
    int b = 20;

      // Function call
    printf("%d\n",__builtin_popcount(a^b));
    return 0;
}

// This code is contributed by Suruchi Kumari

Java

//java code

public class Main {
    public static void main(String[] args) {
        int a = 10;
        int b = 20;

        // Function call
        System.out.println(Integer.bitCount(a ^ b));
    }
}
//code by ksam24000

Python3

# Python program to Count number of bits to be flipped
# to convert A into B

# Driver code
if __name__ == '__main__':
    a = 10
    b = 20

    # Function call
    # Converting int to binary and counting number of bits
    result = bin(a ^ b).count("1")
    print(result)

C#

using System;
class Program
{
  static void Main(string[] args)
  {
    int a = 10;
    int b = 20;

    // Function call
    Console.WriteLine(BitCount(a ^ b));
  }

  static int BitCount(int value)
  {
    int count = 0;
    while (value != 0)
    {
      count += value & 1;
      value >>= 1;
    }
    return count;
  }
}

Javascript

// Function to count number of bits to be flipped
// to convert A into B
function countBits(a, b) {
    return (a ^ b).toString(2).split('1').length-1;
}

// Driver code
let a = 10;
let b = 20;
console.log(countBits(a, b));

// This code is contributed by Potta Lokesh

Output

4

Time Complexity: O(K) where K is the number of bits
Auxiliary Space: O(1) 

Count the number of bits to be flipped to convert A to B using the AND operator:

To solve the problem follow the below idea:

Start comparing the bits in A and B, starting from the least significant bit and if (A & 1) is not equal to (B & 1) then the current bit needs to be flipped, as the value of bits is different at this position in both the numbers

Follow the given steps to solve the problem:

• Declare variable flips equal to zero
• Run a loop, while a is greater than zero and b is also greater than zero
  • Calculate values of (A AND 1) and (B AND 1)
  • If these values are not equal then increase the flip value by 1
  • Right shift a and b by 1
• Return flips

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

int countFlips(int a, int b)
{

    // initially flips is equal to 0
    int flips = 0;

    // & each bits of a && b with 1
    // and store them if t1 and t2
    // if t1 != t2 then we will flip that bit

    while (a > 0 || b > 0) {

        int t1 = (a & 1);
        int t2 = (b & 1);

        if (t1 != t2) {
            flips++;
        }
        // right shifting a and b
        a >>= 1;
        b >>= 1;
    }

    return flips;
}

int main()
{
    int a = 10;
    int b = 20;
    cout << countFlips(a, b);
}

// this code is contributed by shivanisinghss2110

Java

// Java program for the above approach

// CONTRIBUTED BY PRAVEEN VISHWAKARMA

import java.io.*;

class GFG {

    public static int countFlips(int a, int b)
    {
        // initially flips is equal to 0
        int flips = 0;

        // & each bits of a && b with 1
        // and store them if t1 and t2
        // if t1 != t2 then we will flip that bit

        while (a > 0 || b > 0) {

            int t1 = (a & 1);
            int t2 = (b & 1);

            if (t1 != t2) {
                flips++;
            }
            // right shifting a and b
            a >>>= 1;
            b >>>= 1;
        }

        return flips;
    }

    // Driver code
    public static void main(String[] args)
    {
        int a = 10;
        int b = 20;

          // Function call
        System.out.println(countFlips(a, b));
    }
}

Python3

# Python3 program for the above approach


def countFlips(a, b):

    # initially flips is equal to 0
    flips = 0

    # & each bits of a && b with 1
    # and store them if t1 and t2
    # if t1 != t2 then we will flip that bit
    while(a > 0 or b > 0):
        t1 = (a & 1)
        t2 = (b & 1)
        if(t1 != t2):
            flips += 1

        # right shifting a and b
        a >>= 1
        b >>= 1

    return flips


# Driver code
if __name__ == "__main__":
  a = 10
  b = 20

  # Function call
  print(countFlips(a, b))

# This code is contributed by shivanisinghss2110

C#

// C# program for the above approach
using System;

class GFG {

    public static int countFlips(int a, int b)
    {

        // initially flips is equal to 0
        int flips = 0;

        // & each bits of a && b with 1
        // and store them if t1 and t2
        // if t1 != t2 then we will flip that bit
        while (a > 0 || b > 0) {

            int t1 = (a & 1);
            int t2 = (b & 1);

            if (t1 != t2) {
                flips++;
            }
            // right shifting a and b
            a >>= 1;
            b >>= 1;
        }

        return flips;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int a = 10;
        int b = 20;

          // Function call
        Console.Write(countFlips(a, b));
    }
}

// This code is contributed by shivanisinghss2110

Javascript

/*package whatever //do not write package name here */

function countFlips(a, b){
          // initially flips is equal to 0
        var flips = 0;

          // & each bits of a && b with 1
          // and store them if t1 and t2
        // if t1 != t2 then we will flip that bit

          while(a>0 || b>0){

            var t1 = (a&1);
              var t2 = (b&1);

              if(t1!=t2){
                flips++;
            }
              // right shifting a and b
              a>>>=1;
              b>>>=1;
        }

      return flips;
    }

          var a = 10;
          var b = 20;
          document.write(countFlips(a, b));

// This code is contributed by shivanisinghss2110

Output

4

Time Complexity: O(K) where K is the number of bits
Auxiliary Space: O(1)

Count the number of bits to be flipped to convert A to B:-

To solve this we just need to do a few simple steps. To know more follow the below steps:-

Approach:

• Convert A and B to binary numbers.
• Compare using ‘equal to’ operator if equal then return 0 otherwise iterate and 
• compare the ith of A to ith of B and count the operations 
• print the count.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

string binary(int num)
{
    string str = "";
    while (num) {
        if (num & 1) // 1
            str += '1';
        else // 0
            str += '0';
        num >>= 1; // Right Shift by 1
    }
    reverse(str.begin(), str.end());
    return str;
}

int main()
{
    int a = 10;
    int b = 20;
    string astr = binary(a);
    string bstr = binary(b);
    // size of the binary strings.
    int na = astr.size(), nb = bstr.size();
    int cnt = 0;
    // difference between the size of the both a and b
    // string
    //   int maxi = max(na, nb);
    int diff = abs(na - nb);

    // if a size is greater then check it has 1 upto diff
    // then cnt++;

    if (na > nb) {
        for (int i = 0; i < diff; i++) {
            if (astr[i] == '1') {
                cnt++;
            }
        }
    }
    // do the same as above
    else if (na < nb) {
        for (int i = 0; i < diff; i++) {
            if (bstr[i] == '1') {
                cnt++;
            }
        }
    }
    na = na - 1;
    nb = nb - 1;
    // check from the last if has not equal characters and
    // cnt++;
    while (na >= 0 and nb >= 0) {
        if (astr[na] != bstr[nb]) {
            cnt++;
        }
        na--;
        nb--;
    }
    // print the cnt
    cout << cnt << endl;
    return 0;
}

// this code is contributed by ksam24000

Java

import java.util.*;

public class Main {

  public static String binary(int num) {
    String str = "";
    while (num > 0) {
      if ((num & 1) == 1) // 1
        str += '1';
      else // 0
        str += '0';
      num >>= 1; // Right Shift by 1
    }
    return new StringBuilder(str).reverse().toString();
  }

  public static void main(String[] args) {
    int a = 10;
    int b = 20;
    String astr = binary(a);
    String bstr = binary(b);

    // size of the binary strings.
    int na = astr.length(), nb = bstr.length();
    int cnt = 0;

    // difference between the size of the both a and b
    int diff = Math.abs(na - nb);

    // if a size is greater then check it has 1 upto diff
    // then cnt++;
    if (na > nb) {
      for (int i = 0; i < diff; i++) {
        if (astr.charAt(i) == '1') {
          cnt++;
        }
      }
    }

    // do the same as above
    else if (na < nb) {
      for (int i = 0; i < diff; i++) {
        if (bstr.charAt(i) == '1') {
          cnt++;
        }
      }
    }
    na = na - 1;
    nb = nb - 1;

    // check from the last if has not equal characters and
    // cnt++;
    while (na >= 0 && nb >= 0) {
      if (astr.charAt(na) != bstr.charAt(nb)) {
        cnt++;
      }
      na--;
      nb--;
    }

    // print the cnt
    System.out.println(cnt);
  }
}

// This code is contributed by divyansh2212

Python3

def binary(num):
    str = ""
    while num:
        if num & 1: # 1
            str += '1'
        else: # 0
            str += '0'
        num >>= 1 # Right Shift by 1
    return str[::-1]

a = 10
b = 20
astr = binary(a)
bstr = binary(b)

# size of the binary strings.
na, nb = len(astr), len(bstr)
cnt = 0

# difference between the size of the both a and b string
diff = abs(na - nb)

# if a size is greater then check it has 1 upto diff then cnt++
if na > nb:
    for i in range(diff):
        if astr[i] == '1':
            cnt += 1
# do the same as above
elif na < nb:
    for i in range(diff):
        if bstr[i] == '1':
            cnt += 1

na -= 1
nb -= 1
# check from the last if has not equal characters and cnt++
while na >= 0 and nb >= 0:
    if astr[na] != bstr[nb]:
        cnt += 1
    na -= 1
    nb -= 1

# print the cnt
print(cnt)

C#

using System;

public class Program
{
    static string Binary(int num)
    {
        string str = "";
        while (num != 0)
        {
            if ((num & 1) == 1) // 1
                str += '1';
            else // 0
                str += '0';
            num >>= 1; // Right Shift by 1
        }
        char[] charArray = str.ToCharArray();
        Array.Reverse(charArray);
        return new string(charArray);
    }

    public static void Main()
    {
        int a = 10;
        int b = 20;
        string astr = Binary(a);
        string bstr = Binary(b);
        // size of the binary strings.
        int na = astr.Length, nb = bstr.Length;
        int cnt = 0;
        // difference between the size of the both a and b
        // string
        int diff = Math.Abs(na - nb);

        // if a size is greater then check it has 1 upto diff
        // then cnt++;

        if (na > nb)
        {
            for (int i = 0; i < diff; i++)
            {
                if (astr[i] == '1')
                {
                    cnt++;
                }
            }
        }
        // do the same as above
        else if (na < nb)
        {
            for (int i = 0; i < diff; i++)
            {
                if (bstr[i] == '1')
                {
                    cnt++;
                }
            }
        }
        na = na - 1;
        nb = nb - 1;
        // check from the last if has not equal characters and
        // cnt++;
        while (na >= 0 && nb >= 0)
        {
            if (astr[na] != bstr[nb])
            {
                cnt++;
            }
            na--;
            nb--;
        }
        // print the cnt
        Console.WriteLine(cnt);
    }
}

// ksam24000

Javascript

function binary(num) {
  let str = '';
  while (num) {
    if (num & 1) str += '1';
    else str += '0';
    num >>= 1;
  }
  return str.split('').reverse().join('');
}

let a = 10;
let b = 20;
let astr = binary(a);
let bstr = binary(b);
let na = astr.length;
let nb = bstr.length;
let cnt = 0;
let diff = Math.abs(na - nb);

if (na > nb) {
  for (let i = 0; i < diff; i++) {
    if (astr[i] === '1') cnt++;
  }
} else if (na < nb) {
  for (let i = 0; i < diff; i++) {
    if (bstr[i] === '1') cnt++;
  }
}

na--;
nb--;

while (na >= 0 && nb >= 0) {
  if (astr[na] !== bstr[nb]) cnt++;
  na--;
  nb--;
}

console.log(cnt);

// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL

Output

4

Time complexity- O(log N) 
Auxiliary Space – O(N)

Thanks to Sahil Rajput for providing the above implementation.
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