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# Count of possible Strings by replacing consonants with nearest vowel

Given a string str consisting of N letters, the task is to find the total number of strings that can be generated by replacing each consonant with the vowel closest to it in the English alphabet.

Examples:

Input: str = “code”
Output: 2
Explanation: Str = “code” has two consonant c and d.
Closest vowel to d is e and closest to c are a and e.
The possible strings are “aoee” and “eoee

Input: str = “geeks”
Output: 2

Approach: The problem can be solved based on the following observation:

There are total 21 consonant in which ‘c’, ‘g’, ‘l’ and ‘r’ are consonant which is closest to two vowels.
So only these consonants have 2 choices and the remaining have one choices each.
Therefore, the total number of possible strings = the product of the number of choices for each consonant.

Follow the steps mentioned below to implement the observation:

• Initialize a variable (say res = 1) to store the number of possible strings.
• Iterate through the string from i = 0 to N:
• If the character is one of the four special consonants mentioned above then they have two choices. So multiply 2 with res.
• Otherwise, multiply 1 with the value of res.
• The final value of res is the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ code to implement the approach`   `#include ` `using` `namespace` `std;`   `// Function to find the total number of` `// distinct beautiful strings` `int` `uniqueString(string str)` `{` `    ``long` `long` `int` `res = 1;` `    ``for` `(``int` `i = 0; i < str.length(); i++) {` `        ``if` `(str[i] == ``'c'` `|| str[i] == ``'g'` `            ``|| str[i] == ``'l'` `|| str[i] == ``'r'``) {` `            ``res = res * 2;` `        ``}` `    ``}`   `    ``// Return res as which is` `    ``// total number of possible strings` `    ``return` `res;` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``"code"``;`   `    ``// Function call` `    ``cout << (uniqueString(str));` `    ``return` `0;` `}`

## Java

 `// java code to implement the approach`   `import` `java.io.*;`   `class` `GFG {` `    ``// Function to find the total number of distinct beautiful strings` `    ``public` `static` `int` `beautyString(String str)` `    ``{` `        ``char` `alpha[]` `            ``= str.toCharArray(); ``// converting string into` `        ``// array` `        ``int` `res = ``1``, count = ``0``;` `        ``// count for character 'c', 'g', 'l' and 'r'` `        ``for` `(``int` `i = ``0``; i < str.length(); i++) {` `            ``if` `(alpha[i] == ``'c'` `|| alpha[i] == ``'g'` `                ``|| alpha[i] == ``'l'` `|| alpha[i] == ``'r'``) {` `                ``count++;` `                ``res = res * ``2``;` `            ``}` `        ``}` `        ``return` `res;` `    ``}` `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String str = ``"code"``;` `        ``// Function Call` `        ``System.out.println(beautyString(str));` `    ``}` `}`

## Python3

 `# Python code to implement the approach`   `# Function to find the total number of` `# distinct beautiful strings` `def` `uniqueString(s):` `    ``res ``=` `1` `    ``for` `i ``in` `range``(``len``(s)):` `        ``if` `s[i] ``=``=` `'c'` `or` `s[i] ``=``=` `'g'` `or` `s[i] ``=``=` `'l'` `or` `s[i] ``=``=` `'r'``:` `            ``res ``=` `res ``*` `2`   `    ``# Return res as which is` `    ``# total number of possible strings` `    ``return` `res`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` `    ``s ``=` `"code"` `    `  `    ``# Function call` `    ``print``(uniqueString(s))`   `# This code is contributed by Rohit Pradhan`

## C#

 `// C# code to implement the above approach` `using` `System;`   `public` `class` `GFG {`   `  ``// Function to find the total number of distinct beautiful strings` `  ``public` `static` `int` `beautyString(``string` `str)` `  ``{` `    ``char` `[]alpha = str.ToCharArray(); ``// converting string into` `    ``// array` `    ``int` `res = 1, count = 0;`   `    ``// count for character 'c', 'g', 'l' and 'r'` `    ``for` `(``int` `i = 0; i < str.Length; i++) {` `      ``if` `(alpha[i] == ``'c'` `|| alpha[i] == ``'g'` `          ``|| alpha[i] == ``'l'` `|| alpha[i] == ``'r'``) {` `        ``count++;` `        ``res = res * 2;` `      ``}` `    ``}` `    ``return` `res;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``string` `str = ``"code"``;`   `    ``// Function Call` `    ``Console.WriteLine(beautyString(str));` `  ``}` `}`   `// This code is contributed by AnkThon`

## Javascript

 ``

Output

`2`

Time Complexity: O(N)In the above-given approach, there is one loop for iterating over string which takes O(N) time in worst case. Therefore, the time complexity for this approach will be O(N).
Auxiliary Space: O(1) // since no extra array is used so the space taken by the algorithm is constant

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