Rotate bits of a number


Problem link

Code:

package bitManipulation;

public class RotateBits {
    static final int INT_BITS = 32;
   
    static void rotateRight(int n, int k){
        n = (n>>k) | (n<<(INT_BITS-k));
        System.out.println("Right rotaed: "+n);
    }
   
    static void rotateLeft(int n, int k){
        n = (n<<k) | (n>>(INT_BITS-k));
        System.out.println("Left rotaed: "+n);
    }
   
    public static void main(String[] args){
       
        int n = 16, k = 2;
        rotateRight(n,k);
        rotateLeft(n,k);
    }

}




Output:

Right rotaed: 4
Left rotaed: 64

Comments

Post a Comment

Popular posts from this blog

Rearrange Array in Maximum-Minimum form

Given a sorted array of positive integers, rearrange the array alternately i.e first element should be the maximum value, second minimum value, third-second max, fourth-second min and so on. Naive Solution: Here we will use an auxiliary array to store the elements in correct order. For even indices of the array  Large numbers from the given array are added. For odd indices small numbers are added. Code: import java.util.*; public class Test{        public static void main(String[] args) {          Scanner in = new Scanner(System.in);     int[] a = {1,2,3,4,5,6,7,8,9};     int n = a.length;     int[] temp = new int[n];     int small = 0;     int large = n-1;     for(int i=0; i<n; i++){         if(i%2 == 0){             temp[i] = a[large--];         }         else{             temp[i] = a[small++];         }              }     for(int i=0; i<n; i++){         System.out.print(temp[i]+" ");     }            
Read more

Second Largest Element

Given an array of integers, our task is to write a program that efficiently finds the second largest element present in the array. Example: Input : arr[] = {12, 35, 1, 10, 34, 1} Output : The second largest element is 34. Code: package sap; public class SecondLargest {         public static void main(String[] args){                 int[] a = {89, 24, 75, 11, 23};         int max = Integer.MIN_VALUE, secondMax = Integer.MIN_VALUE;                 for(int i=0; i<a.length; i++){                         if(a[i] > max){                                 secondMax = max;                 max = a[i];                             }             else if(a[i] > secondMax){                 secondMax = a[i];             }                     }                 System.out.println("The second maximum element is "+secondMax);             } } Output: The second maximum element is 75
Read more

Check if a number is a power of another number

Check if a number is a power of another number Input: x = 10, y = 1 Output: True x^0 = 1 Input: x = 10, y = 1000 Output: True x^3 = 1 Input: x = 10, y = 1001 Output: False Code: package sap; public class Power {         public static void main(String[] args){                int x = 27, y = 729;                 int val = (int)Math.log(y)/ (int)Math.log(x);                 double res = Math.log(y)/Math.log(x);                 if(val == res){             System.out.println(x+" can be expressed as a power of "+y);         }                 else{             System.out.println(x +" cannot be expressed as a power of "+y);         }             } } Output: 27 can be expressed as a power of 729
Read more