# Reverse actual bits of the given number

• Difficulty Level : Medium
• Last Updated : 03 Aug, 2022

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

```Input : 11
Output : 13
(11)10 = (1011)2.
After reversing the bits we get:
(1101)2 = (13)10.

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

Recommended Practice

In this approach, one by one bit in the binary representation of n is being obtained with the help of bitwise right shift operation and they are being accumulated in rev with the help of bitwise left shift operation.
Algorithm:

## C++

 `// C++ implementation to reverse bits of a number``#include ` `using` `namespace` `std;` `// function to reverse bits of a number``unsigned ``int` `reverseBits(unsigned ``int` `n)``{``    ``unsigned ``int` `rev = 0;` `    ``// traversing bits of 'n' from the right``    ``while` `(n > 0) {``        ``// bitwise left shift``        ``// 'rev' by 1``        ``rev <<= 1;` `        ``// if current bit is '1'``        ``if` `(n & 1 == 1)``            ``rev ^= 1;` `        ``// bitwise right shift``        ``// 'n' by 1``        ``n >>= 1;``    ``}` `    ``// required number``    ``return` `rev;``}` `// Driver program to test above``int` `main()``{``    ``unsigned ``int` `n = 11;``    ``cout << reverseBits(n);``    ``return` `0;``}`

## C

 `// C implementation to reverse bits of a number``#include ` `// function to reverse bits of a number``unsigned ``int` `reverseBits(unsigned ``int` `n)``{``    ``unsigned ``int` `rev = 0;` `    ``// traversing bits of 'n' from the right``    ``while` `(n > 0) {``        ``// bitwise left shift 'rev' by 1``        ``rev <<= 1;` `        ``// if current bit is '1'``        ``if` `(n & 1 == 1)``            ``rev ^= 1;` `        ``// bitwise right shift 'n' by 1``        ``n >>= 1;``    ``}``    ``// required number``    ``return` `rev;``}` `// Driver program to test above``int` `main()``{``    ``unsigned ``int` `n = 11;``    ``printf``(``"%u"``,reverseBits(n));``    ``return` `0;``}` `// This code is contributed by Sania Kumari Gupta (kriSania804)`

## Java

 `// Java implementation to``// reverse bits of a number``class` `GFG``{``    ``// function to reverse bits of a number``    ``public` `static` `int` `reverseBits(``int` `n)``    ``{``        ``int` `rev = ``0``;` `        ``// traversing bits of 'n'``        ``// from the right``        ``while` `(n > ``0``)``        ``{``            ``// bitwise left shift``            ``// 'rev' by 1``            ``rev <<= ``1``;` `            ``// if current bit is '1'``            ``if` `((``int``)(n & ``1``) == ``1``)``                ``rev ^= ``1``;` `            ``// bitwise right shift``            ``//'n' by 1``            ``n >>= ``1``;``        ``}``        ``// required number``        ``return` `rev;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``11``;``        ``System.out.println(reverseBits(n));``    ``}``}` `// This code is contributed``// by prerna saini.`

## Python3

 `# Python 3 implementation to``# reverse bits of a number`  `# function to reverse``# bits of a number``def` `reverseBits(n) :``    ` `    ``rev ``=` `0``    ` `    ``# traversing bits of 'n' from the right``    ``while` `(n > ``0``) :``        ` `        ``# bitwise left shift 'rev' by 1``        ``rev ``=` `rev << ``1``        ` `        ``# if current bit is '1'``        ``if` `(n & ``1` `=``=` `1``) :``            ``rev ``=` `rev ^ ``1``        ` `        ``# bitwise right shift 'n' by 1``        ``n ``=` `n >> ``1``        ` `    ` `    ``# required number``    ``return` `rev``    ` `# Driver code``n ``=` `11``print``(reverseBits(n))`  `# This code is contributed``# by Nikita Tiwari.`

## C#

 `// C# implementation to``// reverse bits of a number``using` `System;``class` `GFG``{``    ``// function to reverse bits of a number``    ``public` `static` `int` `reverseBits(``int` `n)``    ``{``        ``int` `rev = 0;` `        ``// traversing bits of 'n'``        ``// from the right``        ``while` `(n > 0)``        ``{``            ``// bitwise left shift``            ``// 'rev' by 1``            ``rev <<= 1;` `            ``// if current bit is '1'``            ``if` `((``int``)(n & 1) == 1)``                ``rev ^= 1;` `            ``// bitwise right shift``            ``//'n' by 1``            ``n >>= 1;``        ``}``        ``// required number``        ``return` `rev;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 11;``        ``Console.WriteLine(reverseBits(n));``    ``}``}` `// This code is contributed``// by vt_m.`

## PHP

 ` 0)``    ``{``        ``// bitwise left shift``        ``// 'rev' by 1``        ``\$rev` `<<= 1;``        ` `        ``// if current bit is '1'``        ``if` `(``\$n` `& 1 == 1)``            ``\$rev` `^= 1;``        ` `        ``// bitwise right shift``        ``// 'n' by 1``        ``\$n` `>>= 1;``            ` `    ``}``    ` `    ``// required number``    ``return` `\$rev``;``}` `// Driver code``\$n` `= 11;``echo` `reverseBits(``\$n``);` `// This code is contributed by mits``?>`

## Javascript

 ``

Output

`13`

Time Complexity: O(num), where num is the number of bits in the binary representation of n.

Space Complexity: O(1)

Another twist to this problem is to reverse all 4 bytes of an integer value. For e.g. if number is 11 (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1) then its reverse will be -805306368 (1,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0).

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;``import` `java.util.List;``import` `java.util.ArrayList;` `class` `ReverseBits190 {``    ``public` `static` `void` `main (String[] args) {``          ``int` `num = ``11``;``        ``showBits(num); ``// just to show full bit sequence``          ``int` `ret = reverseBits(num);``        ``System.out.println(``"\nreverse of number "` `+ num + ``" is="` `+ ret);``          ``showBits(ret);` `          ``System.out.println(``"\n"``);``        ``num = -``10``;``        ``showBits(num); ``// just to show full bit sequence``          ``ret = reverseBits(num);``        ``System.out.println(``"\nreverse of number "` `+ num + ``" is="` `+ ret);``          ``showBits(ret);``    ``}` `      ``static` `int` `reverseBits(``int` `n) {``        ``int` `newN = ``0``;``        ``for``(``int` `i = ``0``; i < Integer.SIZE; i++) {``            ``newN = newN << ``1``;``            ``if``((n & ``1``) > ``0``) {``                ``newN = newN ^ ``1``;``            ``}``            ``n = n >> ``1``;``        ``}``        ``return` `newN;``    ``}` `  ``// helper method to show actual bits``      ``static` `void` `showBits(``int` `n) {``        ``List l = ``new` `ArrayList<>();``        ``for``(``int` `i = ``0``; i< Integer.SIZE; i++) {``            ` `            ``if``((n & ``1``) > ``0``) l.add(``1``);``            ``else` `l.add(``0``);``            ` `            ``n = n >> ``1``;``        ``}``        ``for``(``int` `i = l.size()-``1``; i >= ``0``; i--) {``            ``System.out.print(l.get(i) + ``","``);``        ``}``    ``}``}`

Output

0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,
reverse of number 11 is=-805306368
1,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,
reverse of number -10 is=1879048191
0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

Contributed by Nikhil.

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