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Invert actual bits of a number
  • 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.
?>
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 number, we have calculated the number of bits in the binary representation, and flipped only the actual bits of 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
?>
Output :
4

Time complexity: O(log n)
Auxiliary space: O(1)

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