# 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
‘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’

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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