# Invert actual bits of a number

• Difficulty Level : Easy

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:

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```Input : 11
Output : 4
(11)10 = (1011)
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 ``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

 ``

## Javascript

 ``
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 ``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

 ``

## Javascript

 ``
Output :
`4`

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

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 ``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;``}`

Output:

`4`

Time complexity: O(1)

Auxiliary space: O(1)

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