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Invert actual bits of a number

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  • Difficulty Level : Easy
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Given a non-negative integer n. The problem is to invert the bits of n and print the number obtained after inverting the bits. Note that the actual binary representation of the number is being considered for inverting the bits, no leading 0’s are being considered. 
 

Examples: 

Input : 11
Output : 4
(11)10 = (1011)[2]
After inverting the bits, we get:
(0100)2 = (4)10.

Input : 10
Output : 5
(10)10 = (1010)2.
After reversing the bits we get:
(0101)2 = (101)2
        = (5)10.

Method 1 (Using bitwise operators)
Prerequisite: Toggling k-th bit of a number 
 

C++




// CPP program to invert actual bits
// of a number.
#include <bits/stdc++.h>
using namespace std;
 
void invertBits(int num)
{
    // calculating number of bits
    // in the number
    int x = log2(num) + 1;
 
    // Inverting the bits one by one
    for (int i = 0; i < x; i++)
       num = (num ^ (1 << i));
  
    cout << num;
}
 
// Driver code
int main()
{
    int num = 11;
    invertBits(num);
    return 0;
}

Java



// Java program to invert
// actual bits of a number.
import java.io.*;
 
class GFG
{
    static void invertBits(int num)
    {
        // calculating number of
        // bits in the number
        int x = (int)(Math.log(num) /
                      Math.log(2)) + 1;
     
        // Inverting the
        // bits one by one
        for (int i = 0; i < x; i++)
        num = (num ^ (1 << i));
     
        System.out.println(num);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int num = 11;
        invertBits(num);
    }
 
}
 
// This code is contributed
// by Anuj_67
Python3



# Python3 program to invert actual
# bits of a number.
import math
 
def invertBits(num):
 
    # calculating number of bits
    # in the number
    x = int(math.log2(num)) + 1
 
    # Inverting the bits one by one
    for i in range(x):
        num = (num ^ (1 << i))
     
    print(num)
 
# Driver Code
num = 11
invertBits(num)
 
# This code is contributed
# by Rituraj Jain
C#



// C# program to invert
// actual bits of a number.
using System;
 
class GFG
{
    static void invertBits(int num)
    {
        // calculating number of
        // bits in the number
        int x = (int)(Math.Log(num) /
                       Math.Log(2)) + 1;
     
        // Inverting the
        // bits one by one
        for (int i = 0; i < x; i++)
        num = (num ^ (1 << i));
     
        Console.WriteLine(num);
    }
     
    // Driver code
    public static void Main ()
    {
        int num = 11;
        invertBits(num);
    }
 
}
 
// This code is contributed
// by Anuj_67
PHP



<?php
// PHP program to invert actual bits
// of a number.
 
function invertBits( $num)
{
     
    // calculating number of bits
    // in the number
    $x = log($num) + 1;
 
    // Inverting the bits one by one
    for($i = 0; $i < $x; $i++)
    $num = ($num ^ (1 << $i));
 
    echo $num;
}
 
    // Driver code
    $num = 11;
    invertBits($num);
     
// This code is contributed by anuj_67.
?>
Javascript



<script>
 
    // Javascript program to invert
    // actual bits of a number.
     
    function invertBits(num)
    {
        // calculating number of
        // bits in the number
        let x =
        parseInt(Math.log(num) / Math.log(2), 10) + 1;
       
        // Inverting the
        // bits one by one
        for (let i = 0; i < x; i++)
            num = (num ^ (1 << i));
       
        document.write(num);
    }
     
    let num = 11;
      invertBits(num);
 
</script>
Output
4
Time complexity : O(log n) 
Auxiliary space : O(1)
Method 2 (Using Bitset) 
Here we use the flip() of bitset to invert the bits of the number, in order to avoid flipping the leading zeroes in the binary representation of the number, we have calculated the number of bits in the binary representation and flipped only the actual bits of the number. We have used to_ulong() to convert bitset to number.
 
C++



// CPP program to invert actual bits
// of a number.
#include <bits/stdc++.h>
using namespace std;
 
void invertBits(int num)
{
    // calculating number of bits
    // in the number
    int x = log2(num) + 1;
 
    // Considering number to be 32 bit integer;
    bitset<32> b(num);
 
    // reversing the bits one by one
    for (int i = 0; i < x; i++)
        b.flip(i);
 
    // converting bitset to number
    cout << b.to_ulong();
}
 
// Driver code
int main()
{
    int num = 11;
    invertBits(num);
    return 0;
}
Java



// Java program to invert actual
// bits of a number.
class GFG
{
    static void invertBits(int num)
    {
        // calculating number of
        // bits in the number
        int x = (int)(Math.log(num) /
                      Math.log(2)) + 1;
     
        // Inverting the bits
        // one by one
         
        for (int i = 0; i < x; i++)
        num = (num ^ (1 << i));
     
        System.out.print(num);
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int num = 11;
        invertBits(num);
    }
}
 
// This code is contributed by Mukul Singh
Python 3



# Python 3 program to invert actual
# bits of a number.
import math
 
def invertBits(num):
     
    # calculating number of
    # bits in the number
    x = int(math.log(num, 2.0) + 1);
 
    # Inverting the bits
    # one by one
    for i in range(0, x):
        num = (num ^ (1 << i));
 
    print(num);
 
# Driver code
num = 11;
invertBits(num);
 
# This code is contributed
# by Akanksha Rai
C#



// C# program to invert actual
// bits of a number.
using System;
 
class GFG
{
    static void invertBits(int num)
    {
        // calculating number of
        // bits in the number
        int x = (int)Math.Log(num, 2) + 1;
     
        // Inverting the bits
        // one by one
        for (int i = 0; i < x; i++)
        num = (num ^ (1 << i));
     
        Console.Write(num);
    }
     
    // Driver code
    static void Main()
    {
        int num = 11;
        invertBits(num);
    }
 
}
 
// This code is contributed by Anuj_67
PHP



<?php
// PHP program to invert actual
// bits of a number.
function invertBits($num)
{
    // calculating number of
    // bits in the number
    $x = log($num) + 1;
 
    // Inverting the bits
    // one by one
    for ($i = 0; $i < $x; $i++)
    $num = ($num ^ (1 << $i));
 
        echo $num;
}
 
// Driver code
$num = 11;
invertBits($num);
 
// This code is contributed
// by ajit
?>
Javascript



<script>
    // Javascript program to invert actual bits of a number.
     
    function invertBits(num)
    {
        // calculating number of
        // bits in the number
        let x = Math.log(num, 2) + 1;
       
        // Inverting the bits
        // one by one
        for (let i = 0; i < x; i++)
        num = (num ^ (1 << i));
       
        document.write(num);
    }
     
    let num = 11;
      invertBits(num);
 
</script>
Output
4
Time complexity: O(log n) 
Auxiliary space: O(1)
Method 3 (Using bitmask):
Here, we will create a mask by setting all the bits from MSB(including MSB) and then take XOR with the original number
C++



// CPP program to invert actual bits
// of a number.
#include <bits/stdc++.h>
using namespace std;
 
void invertBits(int num)
{
    // calculating the mask
    int x = num;     // say num = 100000
      x |= x >> 1;    // 100000 | 010000 = 110000
      x |= x >> 2;    // 110000 | 001100 = 111100
      x |= x >> 4;    // 111100 | 000011 = 111111
      x |= x >> 8;    // 111111 | 000000 = 111111
      x |= x >> 16;    // 111111 | 000000 = 111111
  
    cout << (num ^ x); // 100000 | 111111 = 011111
}
 
// Driver code
int main()
{
    int num = 11;
    invertBits(num);
    return 0;
}
Java



/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
  public static void invertBits(int num) {
    //base case
    if(num == 0)
    {
      System.out.println(1);
      return;
    }
    // calculating the mask
    int x = num;     // say num = 100000
    x |= x >> 1;    // 100000 | 010000 = 110000
    x |= x >> 2;    // 110000 | 001100 = 111100
    x |= x >> 4;    // 111100 | 000011 = 111111
    x |= x >> 8;    // 111111 | 000000 = 111111
    x |= x >> 16;    // 111111 | 000000 = 111111
     
    System.out.println(num ^ x); // 100000 | 111111 = 011111
  }
  // Driver code
    public static void main(String[] args)
    {
        int num = 11;
        invertBits(num);
    }
}
 
// This code is contributed by lapimpale@
C#



// C# code to implement the approach
 
using System;
 
class GFG {
 
    public static void invertBits(int num)
    {
        // base case
        if (num == 0) {
            Console.WriteLine(1);
            return;
        }
        // calculating the mask
        int x = num; // say num = 100000
        x |= x >> 1; // 100000 | 010000 = 110000
        x |= x >> 2; // 110000 | 001100 = 111100
        x |= x >> 4; // 111100 | 000011 = 111111
        x |= x >> 8; // 111111 | 000000 = 111111
        x |= x >> 16; // 111111 | 000000 = 111111
 
        Console.WriteLine(num
                          ^ x); // 100000 | 111111 = 011111
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        int num = 11;
        invertBits(num);
    }
}
 
// This code is contributed by phasing17
Javascript



// JavaScript program to invert actual bits
// of a number.
 
 
function invertBits(num)
{
    // calculating the mask
    let x = num;     // say num = 100000
      x |= x >> 1;    // 100000 | 010000 = 110000
      x |= x >> 2;    // 110000 | 001100 = 111100
      x |= x >> 4;    // 111100 | 000011 = 111111
      x |= x >> 8;    // 111111 | 000000 = 111111
      x |= x >> 16;    // 111111 | 000000 = 111111
  
    console.log(num ^ x); // 100000 | 111111 = 011111
}
 
// Driver code
let num = 11;
invertBits(num);
 
 
// This code is contributed by phasing17
Output
4
Time complexity: O(1) 
Auxiliary space: O(1)
Method 4 (Extracting only the relevant bits using log and XOR)
The inverted number can be efficiently obtained by:
1. Getting the number of bits using log2
2. Taking XOR of the number and 2 numOfBits – 1 
C++



// CPP program to invert actual bits
// of a number.
#include <bits/stdc++.h>
using namespace std;
 
void invertBits(int num)
{
   // Find number of bits in the given integer
   int numOfBits = (int)log2(num) + 1;
 
   // invert the number by taking
   // xor of n and (2 raised to numOfBits) - 1
   cout <<  (((1 << numOfBits) - 1) ^ num);
}
 
// Driver code
int main()
{
    int num = 11;
    //Function Call
    invertBits(num);
    return 0;
}
 
//This code is contributed by phasing17
Output
4
Time complexity: O(1)
Auxiliary space: O(1)
 

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