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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
 

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C++




// C++ code to rotate bits
// of number
#include<iostream>
 
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<<d, last d bits are 0. To
     put first 3 bits of n at
    last, do bitwise or of n<<d
    with 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<stdio.h>
#define INT_BITS 32
 
/*Function to left rotate n by d bits*/
int leftRotate(int n, unsigned int d)
{
   /* In n<<d, last d bits are 0. To put first 3 bits of n at
     last, do bitwise or of n<<d with 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<<d, last d bits are 0.
       To put first 3 bits of n at
       last, do bitwise or of n<<d with
       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<<d, last d bits are 0.
    # To put first 3 bits of n at
    # last, do bitwise or of n<<d
    # with 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<<d, last d bits are 0.
        To put first 3 bits of n at
        last, do bitwise or of n<<d with
        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




<script>
 
// JavaScript code to rotate bits
// of number
let INT_BITS = 32;
 
     
/*Function to left rotate n by d bits*/
 
function leftRotate( n,  d)
{   
    /* In n<<d, last d bits are 0. To
     put first 3 bits of n at
    last, do bitwise or of n<<d
    with n >>(INT_BITS - d) */
    return (n << d)|(n >> (INT_BITS - d));
}
 
/*Function to right rotate n by d bits*/
function 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));
}
 
 
/* Driver code*/
let n = 16;
let d = 2;
document.write("Left Rotation of " + n +
            " by " + d + " is ");
document.write(leftRotate(n, d));
document.write("<br>");           
document.write("\nRight Rotation of " + n +
            " by " + d + " is ");
document.write(rightRotate(n, d));
 
</script>

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 <bits/stdc++.h>
 
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 <stdio.h>
#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<<D) | 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<<D) | 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




<script>
 
function rotate(N,D)
{
    // your code here
        let t = 16;
        let left = ((N<<D) | N>>(t-D)) & 0xFFFF;
        let right = ((N>>D) | N<<(t-D)) & 0xFFFF;
        document.write(left + "<br>");
         document.write(right + "<br>");
}
 
let N = 28;
let D = 2;
rotate(N, D);
 
// This code is contributed by avanitrachhadiya2155
</script>
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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