# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 // 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));     } }

## Python

 # Python3 implementation of the above approach: MAX = 26    # Function to return the encrypted strring def encryptstrr(strr, n, x):            # Reduce x because rotation of     # length 26 is unnecessary     x = x % MAX     arr = list(strr)            # calculate the frequency of characters     freq = [0]*MAX     for i in range(n):         freq[ord(arr[i]) - ord('a')] += 1            for i in range(n):                    # If the frequency of current character         # is even then increment it by x         if (freq[ord(arr[i]) - ord('a')] % 2 == 0):             pos = (ord(arr[i]) - ord('a') + x) % MAX             arr[i] = chr(pos + ord('a'))                    # Else decrement it by x         else:             pos = (ord(arr[i]) - ord('a') - x)             if (pos < 0):                 pos += MAX             arr[i] = chr(pos + ord('a'))                    # Return the count     return "".join(arr)       # Driver code s = "abcda" n = len(s) x = 3 print(encryptstrr(s, n, x))    # This code is contributed by  # shubhamsingh10

## C#

 // 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

Output:

dyzad

Time Complexity: O(N)

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