Java Program to Rotate bits of a number
Bit Rotation: A rotation (or circular shift) is an operation similar to a shift except that the bits that fall off at one end are put back to the other end.
In the left rotation, the bits that fall off at the left end are put back at the right end.
In the right rotation, the bits that fall off at the right end are put back at the left end.
Example of Rotating bits of a Number
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.
Program to Rotate Bits of a Number in Java
Below is the implementation of Bit rotation in 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) */
d %= INT_BITS;
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) */
d %= INT_BITS;
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.println(leftRotate(n, d));
System.out.print("Right Rotation of " + n +
" by " + d + " is ");
System.out.println(rightRotate(n, d));
}
}
Output :
Left Rotation of 16 by 2 is 64
Right Rotation of 16 by 2 is 4
Complexity of the above method:
Time Complexity: O(1)
Auxiliary Space: O(1)
Method for 16 bit Number
Below is the implementation of the above method:
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;
D %= 16;
// In case, D > 16, t - D will be negative.
int left = ((N << D) | N >> (t - D)) & 0xFFFF;
int right = ((N >> D) | N << (t - D)) & 0xFFFF;
System.out.println("Left Rotation of " + N +
" by " + D + " is "+left);
System.out.println("Right Rotation of " + N +
" by " + D + " is "+right);
}
}
Output:
Left Rotation of 28 by 2 is 112
Right Rotation of 28 by 2 is 7
Complexity of the above Method:
Time Complexity : O(1)
Auxiliary Space: O(1)
Please refer complete article on Rotate bits of a number for more details!
