# Rotate bits of a number

• Difficulty Level : Easy
• Last Updated : 01 Nov, 2021

Bit Rotation: A rotation (or circular shift) is an operation similar to shift except that the bits that fall off at one end are put back to the other end.
In left rotation, the bits that fall off at left end are put back at right end.
In right rotation, the bits that fall off at right end are put back at left end.

Example:
Let n is stored using 8 bits. Left rotation of n = 11100101 by 3 makes n = 00101111 (Left shifted by 3 and first 3 bits are put back in last ). If n is stored using 16 bits or 32 bits then left rotation of n (000…11100101) becomes 00..0011100101000.
Right rotation of n = 11100101 by 3 makes n = 10111100 (Right shifted by 3 and last 3 bits are put back in first ) if n is stored using 8 bits. If n is stored using 16 bits or 32 bits then right rotation of n (000…11100101) by 3 becomes 101000..0011100

## C++

 `// C++ code to rotate bits``// of number``#include` `using` `namespace` `std;``#define INT_BITS 32``class` `gfg``{``    ` `/*Function to left rotate n by d bits*/``public``:``int` `leftRotate(``int` `n, unsigned ``int` `d)``{``    ` `    ``/* In n<>(INT_BITS - d) */``    ``return` `(n << d)|(n >> (INT_BITS - d));``}` `/*Function to right rotate n by d bits*/``int` `rightRotate(``int` `n, unsigned ``int` `d)``{``    ``/* In n>>d, first d bits are 0.``    ``To put last 3 bits of at``    ``first, do bitwise or of n>>d``    ``with n <<(INT_BITS - d) */``    ``return` `(n >> d)|(n << (INT_BITS - d));``}``};` `/* Driver code*/``int` `main()``{``    ``gfg g;``    ``int` `n = 16;``    ``int` `d = 2;``    ``cout << ``"Left Rotation of "` `<< n <<``            ``" by "` `<< d << ``" is "``;``    ``cout << g.leftRotate(n, d);``    ``cout << ``"\nRight Rotation of "` `<< n <<``            ``" by "` `<< d << ``" is "``;``    ``cout << g.rightRotate(n, d);``    ``getchar``();``}` `// This code is contributed by SoM15242`

## C

 `#include``#define INT_BITS 32` `/*Function to left rotate n by d bits*/``int` `leftRotate(``int` `n, unsigned ``int` `d)``{``   ``/* In n<>(INT_BITS - d) */``   ``return` `(n << d)|(n >> (INT_BITS - d));``}` `/*Function to right rotate n by d bits*/``int` `rightRotate(``int` `n, unsigned ``int` `d)``{``   ``/* In n>>d, first d bits are 0. To put last 3 bits of at``     ``first, do bitwise or of n>>d with n <<(INT_BITS - d) */``   ``return` `(n >> d)|(n << (INT_BITS - d));``}` `/* Driver program to test above functions */``int` `main()``{``  ``int` `n = 16;``  ``int` `d = 2;``  ``printf``(``"Left Rotation of %d by %d is "``, n, d);``  ``printf``(``"%d"``, leftRotate(n, d));``  ``printf``(``"\nRight Rotation of %d by %d is "``, n, d);``  ``printf``(``"%d"``, rightRotate(n, d));``  ``getchar``();``}`

## Java

 `// Java code to rotate bits``// of number``class` `GFG``{``static` `final` `int` `INT_BITS = ``32``;` `/*Function to left rotate n by d bits*/``static` `int` `leftRotate(``int` `n, ``int` `d) {``    ` `    ``/* In n<>(INT_BITS - d) */``    ``return` `(n << d) | (n >> (INT_BITS - d));``}` `/*Function to right rotate n by d bits*/``static` `int` `rightRotate(``int` `n, ``int` `d) {``    ` `    ``/* In n>>d, first d bits are 0.``       ``To put last 3 bits of at``       ``first, do bitwise or of n>>d``       ``with n <<(INT_BITS - d) */``    ``return` `(n >> d) | (n << (INT_BITS - d));``}` `// Driver code``public` `static` `void` `main(String arg[])``{``    ``int` `n = ``16``;``    ``int` `d = ``2``;``    ``System.out.print(``"Left Rotation of "` `+ n +``                          ``" by "` `+ d + ``" is "``);``    ``System.out.print(leftRotate(n, d));``    ` `    ``System.out.print(``"\nRight Rotation of "` `+ n +``                             ``" by "` `+ d + ``" is "``);``    ``System.out.print(rightRotate(n, d));``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 code to``# rotate bits of number` `INT_BITS ``=` `32` `# Function to left``# rotate n by d bits``def` `leftRotate(n, d):` `    ``# In n<>(INT_BITS - d)``    ``return` `(n << d)|(n >> (INT_BITS ``-` `d))` `# Function to right``# rotate n by d bits``def` `rightRotate(n, d):` `    ``# In n>>d, first d bits are 0.``    ``# To put last 3 bits of at``    ``# first, do bitwise or of n>>d``    ``# with n <<(INT_BITS - d)``    ``return` `(n >> d)|(n << (INT_BITS ``-` `d)) & ``0xFFFFFFFF` `# Driver program to``# test above functions``n ``=` `16``d ``=` `2` `print``(``"Left Rotation of"``,n,``"by"``      ``,d,``"is"``,end``=``" "``)``print``(leftRotate(n, d))` `print``(``"Right Rotation of"``,n,``"by"``     ``,d,``"is"``,end``=``" "``)``print``(rightRotate(n, d))` `# This code is contributed by``# Smitha Dinesh Semwal`

## C#

 `// C# program to rotate``// bits of a number``using` `System;` `class` `GFG``{``    ``static` `int` `INT_BITS = 32;` `    ``/* Function to left rotate n by d bits*/``    ``static` `int` `leftRotate(``int` `n, ``int` `d) {``        ` `        ``/* In n<>(INT_BITS - d) */``        ``return` `(n << d) | (n >> (INT_BITS - d));``    ``}``    ` `    ``/*Function to right rotate n by d bits*/``    ``static` `int` `rightRotate(``int` `n, ``int` `d) {``        ` `        ``/* In n>>d, first d bits are 0.``        ``To put last 3 bits of at``        ``first, do bitwise or of n>>d``        ``with n <<(INT_BITS - d) */``        ``return` `(n >> d) | (n << (INT_BITS - d));``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 16;``        ``int` `d = 2;``        ` `        ``Console.Write(``"Left Rotation of "` `+ n``                      ``+ ``" by "` `+ d + ``" is "``);``        ``Console.Write(leftRotate(n, d));``        ` `        ``Console.Write(``"\nRight Rotation of "` `+ n``                       ``+ ``" by "` `+ d + ``" is "``);``        ``Console.Write(rightRotate(n, d));``    ``}``}` `// This code is contributed by Sam007`

## Javascript

 ``

Output :

```Left Rotation of 16 by 2 is 64
Right Rotation of 16 by 2 is 4```

Time Complexity: O(1)

Auxiliary Space: O(1)

For 16 bit number:

## C++

 `#include ` `using` `namespace` `std;` `#define SHORT_SIZE 16` `// function to rotate the given unsigned short``// in the left direction``unsigned ``short` `leftRotate(unsigned ``short` `x, ``short` `d)``{``    ``/**``     ``* By doing x << d, we move the first(right most) d bits``     ``* to the left most d bits, and at the same time we move``     ``* the left most d bits to the right side,``     ``* performing OR operation between the two gives use the``     ``* required result.``     ``* */` `    ``return` `(x << d) | (x >> (SHORT_SIZE - d));``}` `// function to rotate the given unsigned short``// in the right direction``unsigned ``short` `rightRotate(unsigned ``short` `x, ``short` `d)``{``    ``/**``     ``* By doing x >> d, we move the first(left most) d bits``     ``* to the right most d bits, and at the same time we move``     ``* the right most d bits to the right side,``     ``* performing OR operation between the two gives use the``     ``* required result.``     ``* */` `    ``return` `(x >> d) | (x << (SHORT_SIZE - d));``}` `/* Driver program to test above functions */``int` `main()``{``    ``// Test case``    ``unsigned ``short` `n = 28;``    ``short` `d = 2;` `    ``cout << leftRotate(n, d) << endl;``    ``cout << rightRotate(n, d) << endl;` `    ``return` `0;``}` `// This code is contributed by ganesh227`

## C

 `#include ``#define SHORT_SIZE 16` `// function to rotate the given unsigned short``// in the left direction``unsigned ``short` `leftRotate(unsigned ``short` `x, ``short` `d)``{``    ``/**``     ``* By doing x << d, we move the first(right most) d bits``     ``* to the left most d bits, and at the same time we move``     ``* the left most d bits to the right side,``     ``* performing OR operation between the two gives use the``     ``* required result.``     ``* */` `    ``return` `(x << d) | (x >> (SHORT_SIZE - d));``}` `// function to rotate the given unsigned short``// in the right direction``unsigned ``short` `rightRotate(unsigned ``short` `x, ``short` `d)``{``    ``/**``     ``* By doing x >> d, we move the first(left most) d bits``     ``* to the right most d bits, and at the same time we move``     ``* the right most d bits to the right side,``     ``* performing OR operation between the two gives use the``     ``* required result.``     ``* */` `    ``return` `(x >> d) | (x << (SHORT_SIZE - d));``}` `/* Driver program to test above functions */``int` `main()``{``    ``// Test case``    ``unsigned ``short` `n = 28;``    ``short` `d = 2;``    ``printf``(``"%d\n"``, leftRotate(n, d));``    ``printf``(``"%d\n"``, rightRotate(n, d));` `    ``return` `0;``}` `// This code is contributed by ganesh227`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `class` `GFG {``    ``public` `static` `void` `main (String[] args) {``        ``int` `N=``28``;``          ``int` `D=``2``;``          ``rotate(N,D);``    ``}``  ` `   ``static` `void` `rotate(``int` `N, ``int` `D)``    ``{``        ``// your code here``        ``int` `t=``16``;``        ``int` `left= ((N<>(t-D)) & ``0xFFFF``;``        ``int` `right=((N>>D) | N<<(t-D)) & ``0xFFFF``;``        ``System.out.println(left);``         ``System.out.println(right);``    ``}``  ` `}`

## Python3

 `SHORT_SIZE ``=` `16` `# function to rotate the given unsigned short``# in the left direction``def` `leftRotate(x, d):` `    ``return` `(x << d) | (x >> (SHORT_SIZE ``-` `d))` `  ``# function to rotate the given unsigned short``# in the right direction``def` `rightRotate(x, d):` `    ``return` `(x >> d) | (x << (SHORT_SIZE ``-` `d)) & ``0xDDDDDF` `# Driver program to test above functions``# Test case``n ``=` `28``d ``=` `2` `print``(``"Left Rotation of"``,n,``"by"``      ``,d,``"is"``,end``=``" "``)``print``(leftRotate(n, d))` `print``(``"Right Rotation of"``,n,``"by"``     ``,d,``"is"``,end``=``" "``)``print``(rightRotate(n, d))` `# This code is contributed by shivanisinghss2110`

## C#

 `/*package whatever //do not write package name here */``using` `System;``using` `System.Collections.Generic;``public` `class` `GFG {``    ``public` `static` `void` `Main(String[] args) {``        ``int` `N = 28;``          ``int` `D = 2;``          ``rotate(N, D);``    ``}``  ` `  ``// Driver code``   ``static` `void` `rotate(``int` `N, ``int` `D)``    ``{``        ``int` `t = 16;``        ``int` `left = ((N<>(t-D)) & 0xFFFF;``        ``int` `right = ((N>>D) | N<<(t-D)) & 0xFFFF;``        ``Console.WriteLine(left);``         ``Console.WriteLine(right);``    ``}``  ` `}` `// This code is contributed by umadevi9616`

## Javascript

 ``

```Left Rotation of 28 by 2 is 112
Right Rotation of 28 by 2 is 7```

Time Complexity : O(1)

Space Complexity : O(1)

Please write comments if you find any bug in the above program or other ways to solve the same problem.

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