Invert actual bits of a number
Last Updated :
11 Jul, 2025
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 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 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# 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
<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>
<?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.
?>
Time complexity : O(log n)
Auxiliary space : O(1)
Method 2 (Using bit-wise operators)
Think about generating a number just less than or equal to 'n' with all bits '1' and take xor of that number with 'n'.
Algorithm:
1. A variable 'x' is initialized to 1.
2. The function enters a while loop that continues until 'x' is less than or equal to the input integer (A).
3. Inside the while loop, 'x' is left-shifted by 1 using the bit-wise left shift operator (<<=). This multiplies 'x' by 2.
4. After the while loop, the value of 'x' is decremented by 1.
5. The function returns the result of the bit-wise XOR operation between 'x' and the input integer (A)
6. The final result is the bit-wise XOR of the input integer with the largest power of 2 that is less than or equal to the input integer.
Below is the implementation of the above idea:
C++
#include <bits/stdc++.h>
using namespace std;
class GFG {
public:
int Opposite(int n)
{
int x = 1;
while (x <= n) {
x <<= 1;
}
x--;
return (x ^ n);
}
};
int main()
{
int n = 5;
GFG Obj = GFG();
cout << Obj.Opposite(n);
return 0;
}
import java.util.*;
class GFG {
public int Opposite(int n) {
int x = 1;
while(x <= n){
x <<= 1;
}
x--;
return (x ^ n);
}
}
public class Main {
public static void main(String[] args) {
int n = 5;
GFG Obj = new GFG();
System.out.print(Obj.Opposite(n));
}
}
class GFG:
def Opposite(self, n):
x = 1
while x <= n:
x <<= 1
x -= 1
return (x ^ n)
n = 5
Obj = GFG()
print(Obj.Opposite(n))
using System;
public class GFG {
public int Opposite(int n)
{
int x = 1;
while (x <= n) {
x <<= 1;
}
x--;
return (x ^ n);
}
}
class Program {
static void Main(string[] args)
{
int n = 5;
GFG Obj = new GFG();
Console.WriteLine(Obj.Opposite(n));
}
}
// Javascript code addition
// Creating a class name GFG
class GFG {
Opposite(n) {
let x = 1;
// Keep multiplying x by 2, until x <= n
while (x <= n) {
x <<= 1;
}
// decrease the value of x by 1.
x--;
// return xor of x and n.
return (x ^ n);
}
}
// assigning value to n
let n = 5;
// creating an object.
let Obj = new GFG();
// calling Opposite() function from the class.
console.log(Obj.Opposite(n));
// The code is contributed by Arushi Goel.
Time Complexity: O(log n)
Auxiliary Space: O(1)
Method 3 (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 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 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# 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
<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>
<?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
?>
Time complexity: O(log n)
Auxiliary space: O(1)
Method 4 (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;
}
/*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@
def invertBits(num):
# calculating the mask
x = num
x |= x >> 1
x |= x >> 2
x |= x >> 4
x |= x >> 8
x |= x >> 16
print(num ^ x)
# Driver Code
num = 11
invertBits(num)
// 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 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
Time complexity: O(1)
Auxiliary space: O(1)
Method 5 (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
// Java program to invert actual bits
// of a number.
import java.io.*;
class GFG {
static void invertBits(int num)
{
// Find number of bits in the given integer
int numOfBits = (int)(Math.log(num) / Math.log(2)) + 1;
// invert the number by taking
// xor of n and (2 raised to numOfBits) - 1
System.out.println(((1 << numOfBits) - 1) ^ num);
}
// Driver code
public static void main(String[] args)
{
int num = 11;
//Function Call
invertBits(num);
}
}
# Python program to invert actual bits
# of a number.
def invertBits(num):
# Find number of bits in the given integer
numOfBits = num.bit_length()
# Invert the number by taking
# xor of n and (2 raised to numOfBits) - 1
print(((1 << numOfBits) - 1) ^ num)
# Driver code
num = 11
# Function Call
invertBits(num)
// C# program to invert actual bits
// of a number.
using System;
class GFG
{
static void invertBits(int num)
{
// Find number of bits in the given integer
int numOfBits = (int)(Math.Log(num) / Math.Log(2)) + 1;
// invert the number by taking
// xor of n and (2 raised to numOfBits) - 1
Console.WriteLine(((1 << numOfBits) - 1) ^ num);
}
// Driver code
public static void Main(string[] args)
{
int num = 11;
// Function Call
invertBits(num);
}
}
// contributed by phasing17
// JavaScript program to invert actual bits
// of a number.
function invertBits(num) {
// Find number of bits in the given integer
let numOfBits = Math.floor(Math.log2(num)) + 1;
// Invert the number by taking
// xor of n and (2 raised to numOfBits) - 1
console.log(((1 << numOfBits) - 1) ^ num);
}
// Driver code
let num = 11;
// Function Call
invertBits(num);
Time complexity: O(1)
Auxiliary space: O(1)
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