Program to invert bits of a number Efficiently

Given a non-negative integer N. The task is to invert the bits of the number N and print the decimal equivalent of the number obtained after inverting the bits. 
Note: Leading 0’s are not being considered.
Examples:

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

Input : 20
Output : 11
(20)10 = (10100)2.
After inverting the bits, we get:
(01011)2 = (11)10.

A similar problem is already discussed in Invert actual bits of a number.
In this article, an efficient approach using bitwise operators is discussed. Below is the step by step algorithm to solve the problem:

1. Calculate the total number of bits in the given number. This can be done by calculating:

   X = log2N

   Where N is the given number and X is the total number of bits of N.

2. The next step is to generate a number with X bits and all bits set. That is, 11111….X-times. This can be done by calculating:

   Step-1: M = 1 << X
   Step-2: M = M | (M-1)

   Where M is the required X-bit number with all bits set.

3. The final step is to calculate the bit-wise XOR of M with N, which will be our answer.

Below is the implementation of the above approach: 

C++

// C++ program to invert actual bits
// of a number.
#include <bits/stdc++.h>

using namespace std;

// Function to invert bits of a number
int invertBits(int n)
{
    // Calculate number of bits of N-1;
    int x = log2(n) ;

    int m = 1 << x;

    m = m | m - 1;

    n = n ^ m;

    return n;
}

// Driver code
int main()
{
    int n = 20;

    cout << invertBits(n);

    return 0;
}

Java

// Java program to invert
// actual bits of a number.
import java.util.*;

class GFG
{
// Function to invert
// bits of a number
static int invertBits(int n)
{
    // Calculate number of bits of N-1;
    int x = (int)(Math.log(n) /
                  Math.log(2)) ;

    int m = 1 << x;

    m = m | m - 1;

    n = n ^ m;

    return n;
}

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

    System.out.print(invertBits(n));
}
}

// This code is contributed by Smitha

Python3

# Python3 program to invert actual
# bits of a number.
import math

# Function to invert bits of a number
def invertBits(n):

    # Calculate number of bits of N-1
    x = int(math.log(n, 2))

    m = 1 << x

    m = m | m - 1

    n = n ^ m

    return n

# Driver code
n = 20

print(invertBits(n))

# This code is contributed 29AjayKumar

C#

// C# program to invert
// actual bits of a number.
using System;



public class GFG
{
// Function to invert
// bits of a number
static int invertBits(int n)
{
    // Calculate number of bits of N-1;
    int x = (int)(Math.Log(n) /
                  Math.Log(2)) ;

    int m = 1 << x;

    m = m | m - 1;

    n = n ^ m;

    return n;
}

// Driver code
public static void Main()
{
    int n = 20;

    Console.Write(invertBits(n));
}
}

// This code is contributed by Subhadeep

PHP

<?php
// PHP program to invert actual
// bits of a number.

// Function to invert bits
// of a number
function invertBits($n)
{
    // Calculate number of
    // bits of N-1;
    $x = log($n, 2);

    $m = 1 << $x;

    $m = $m | $m - 1;

    $n = $n ^ $m;

    return $n;
}

// Driver code
$n = 20;

echo(invertBits($n));

// This code is contributed
// by mits
?>

Javascript

<script>

// Javascript program to invert actual bits
// of a number.

// Function to invert bits of a number
function invertBits(n)
{
    // Calculate number of bits of N-1;
    let x = parseInt(Math.log(n) / Math.log(2)) ;

    let m = 1 << x;

    m = m | m - 1;

    n = n ^ m;

    return n;
}

// Driver code
    let n = 20;

    document.write(invertBits(n));

</script>

Output:

11

Time Complexity: O(log₂n) 
Auxiliary Space: O(1)