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# Depend of attainable Strings by changing 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 reamining 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++

 #include using namespace std;    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; }    int main() {     string str = "code";             cout << (uniqueString(str));     return 0; }

## Java

 import java.io.*;    class GFG {          public static int beautyString(String str)     {         char alpha[]             = str.toCharArray();                  int res = 1, count = 0;                  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;     }          public static void main(String[] args)     {         String str = "code";                  System.out.println(beautyString(str));     } }

Time Complexity: O(N)
Auxiliary Space: O(1)

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