Encrypt the given string with the following operations

Given a string s, the task is to encrypt the string in the following way:

  1. If the frequency of current character is even, then increment current character by x.
  2. If the frequency of current character is odd, then decrement current character by x.

Note: All the operations are circular that is adding 1 to ‘z’ will give ‘a’ and subtracting 1 from ‘a’ will give ‘z’

Examples:



Input :s=”abcda”, x=3
Output :dyzad
‘a’ appear 2 times in the string, hence incrementing ‘a’ by 3 becomes ‘d’
‘b’ appear 1 times in the string, hence decrementing ‘b’ by 3 becomes ‘y’
‘c’ appear 1 times in the string, hence decrementing ‘c’ by 3 becomes ‘z’
‘d’ appear 1 times in the string, hence decrementing ‘d’ by 3 becomes ‘a’
‘a’ appear 2 times in the string, hence incrementing ‘a’ by 3 becomes ‘d’
Hence the string becomes “dyzad”

Input :s=”aabbc”, x=5
Output :ffggx

Approach:

  • Create a frequency array to store the frequency of each character .
  • Traverse the given string and for every character if its frequency is even, then increment it by x else if the frequency is odd decrement it by x.

Below is the implementation of the above approach:

Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the above approach:
public class GFG {
  
    static final int MAX = 26;
  
    // Function to return the encrypted string
    static String encryptStr(String str, int n, int x)
    {
  
        // Reduce x because rotation of
        // length 26 is unnecessary
        x = x % MAX;
        char arr[] = str.toCharArray();
  
        // calculate the frequency of characters
        int freq[] = new int[MAX];
        for (int i = 0; i < n; i++)
            freq[arr[i] - 'a']++;
  
        for (int i = 0; i < n; i++) {
  
            // If the frequency of current character
            // is even then increment it by x
            if (freq[arr[i] - 'a'] % 2 == 0) {
                int pos = (arr[i] - 'a' + x) % MAX;
                arr[i] = (char)(pos + 'a');
            }
  
            // Else decrement it by x
            else {
                int pos = (arr[i] - 'a' - x);
                if (pos < 0)
                    pos += MAX;
                arr[i] = (char)(pos + 'a');
            }
        }
  
        // Return the count
        return String.valueOf(arr);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String s = "abcda";
        int n = s.length();
        int x = 3;
        System.out.println(encryptStr(s, n, x));
    }
}

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach: 
using System;
  
class GFG
{
  
    static int MAX = 26;
  
    // Function to return the encrypted string 
    public static char[] encryptStr(String str, 
                                    int n, int x)
    {
  
        // Reduce x because rotation of 
        // length 26 is unnecessary 
        x = x % MAX;
        char[] arr = str.ToCharArray();
  
        // calculate the frequency of characters 
        int[] freq = new int[MAX];
        for (int i = 0; i < n; i++)
            freq[arr[i] - 'a']++;
  
        for (int i = 0; i < n; i++)
        {
  
            // If the frequency of current character 
            // is even then increment it by x 
            if (freq[arr[i] - 'a'] % 2 == 0)
            {
                int pos = (arr[i] - 'a' + x) % MAX;
                arr[i] = (char)(pos + 'a');
            }
  
            // Else decrement it by x 
            else
            {
                int pos = (arr[i] - 'a' - x);
                if (pos < 0)
                    pos += MAX;
                arr[i] = (char)(pos + 'a');
            }
        }
  
        // Return the count 
        return arr;
    }
  
    // Driver code 
    public static void Main(String[] args)
    {
        String s = "abcda";
        int n = s.Length;
        int x = 3;
        Console.WriteLine(encryptStr(s, n, x));
    }
}
  
// This code is contributed by
// sanjeev2552

chevron_right


Output:

dyzad

Time Complexity: O(N)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : sanjeev2552



Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.