Question

The company Digital Secure Data
Transfer Solutions provides data encryption
and data sharing services. The process uses a
key for encryption when transmitting a list of
elements. To encrypt the list of elements, each
element in the list is replaced by the sum of
odd elements from the next/previous K
elements in the list. If the value of K is positive
then the previous K elements in the list are
selected. If the value of K is negative, then the
next Kelements in the list are selected. The
series of elements is considered in a cyclic
fashion for the last Kelements. If no odd
element in the next/previous k elements is
found then the element is replaced with a zero.
Write an algorithm to find the encrypted list.​

Answer 1

Answer:

def list_encrypter(list_n):

   encrypted_array  =[]

   for i in range(len(list_n)):

       if list_n[i] % 2 == 0:

           encrypted_array.append(sum(list_n[:int(i)]))

       else:

           encrypted_array.append(sum(list_n[int(i):]))

   return encrypted_array

Explanation:

needs no explanation

Answer 2

JAVA CODE

package arrayproblems;

import java.util.ArrayList;

class Problem21 {

   public static void main(String[] args) {

       int N=25143;

       int key=3;

       String res="";

       ArrayList array1=new ArrayList();

       

       while(N!=0) {

           array1.add(N%10);

           N=N/10;

       }

       

       int[] array2=new int[array1.size()];

       int k=0;

       for(int i=array1.size()-1;i>=0;i--) {  

           array2[k]=(int) array1.get(i);

           k=k+1;

       }

       for(int j=0;j<key-1;j++) {

           int temp=array2[array2.length-1];

            for(int i=array2.length-1;i>=1;i--) {        

               array2[i]=array2[i-1];

           }

           array2[0]=temp;

       }

        for(int i:array2) {

           System.out.print(i+res);

       }

   }

}

#SPJ3