Question

write a program that forms a new sentence on taking an input string as per below algorithm the alphabet should be N number of times in forward mode
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Answer 1

Given an array of strings (all lowercase letters), the task is to group them in such a way that all strings in a group are shifted versions of each other. Two string S and T are called shifted if,

S.length = T.length

and

S[i] = T[i] + K for

1 <= i <= S.length for a constant integer K

For example strings, {acd, dfg, wyz, yab, mop} are shifted versions of each other.

Input : str[] = {"acd", "dfg", "wyz", "yab", "mop",

"bdfh", "a", "x", "moqs"};

Output : a x

acd dfg wyz yab mop

bdfh moqs

All shifted strings are grouped together.

Answer 2

Answer:

From the end ($n-1$, where n is the length of an array) to the beginning, traverse the Shift array (index $0$). Take the modulus of $26$ and multiply the value at the previous index by the value at the current index (Total number of characters in English).

Explanation:

The steps to solving this problem are as follows:-

• This step is carried out to determine the total number of shift operations that must be performed on each character of the input string.

• Create a character array from the input string.

• From the beginning (index $0$) to the end (index n), traverse the String array ($n-1$, where n is the length of an array). Apply the shift operation to every single character in an array.

• Return a modified array of characters as a string.

Program:

public class Main {

   public static void main(String[] args) {        

       Main main = new Main();

       String result = main.shiftingLetters("abcd", new int[] {3, 5, 9, 1});

       System.out.print(result);

   }

   

   /* Solution */

   public String shiftingLetters(String S, int[] shifts) {

       

       int previous = 0;

       for(int i=shifts.length-1; i>=0; i--){            

           shifts[i] = (shifts[i] + previous) % 26;

           previous = shifts[i];

       }

       

       char[] chars = S.toCharArray();

       for(int i=0; i<chars.length; i++){

           chars[i] = (char)('a' + (((int)chars[i] + shifts[i]) % 'a') % 26);

       }

       return String.valueOf(chars);

   }

}