Count strings with consonants and vowels at alternate position

Given a string str. The task is to find all possible number of strings that can be obtained by replacing the “$” with alphabets in the given string.

Note: Alphabets should be placed in such a way that the string is always alternating in vowels and consonants, and the string must always start with a consonant. It is assumed that such a string is always possible, i.e. there is no need to care about the characters other than “$”.

Examples: 

Input: str = "y$s"
Output: 5
$ can be replaced with any of the 5 vowels.
So, there can be 5 strings.

Input: str = "s$$e$"
Output 2205

Approach: It is given that the string will start with a consonant. So, if ‘$’ is at even position(considering 0-based indexing) then there should be a consonant else there should be a vowel. Also, given that there is no need to care about the characters other than “$”, i.e., characters other than “$” are placed correctly in the string maintaining the alternating consonant and vowel sequence. Let us understand the problem with an example.

str = “s$$e$”
Here we have to find the number of ways to form a string with given constraints. 

• First occurrence of $ is at 2nd position i.e. 1st index, so we can use 5 vowels.
• Second occurrence of $ is at 3rd position, so we can use 21 consonants.
• Third occurrence of $ is at 5th position, so we can use 21 consonants.

So, total number of ways to form above string is = 5*21*21 = 2205

Below is the implementation of the above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>

using namespace std;

// Function to find the count of strings
int countStrings(string s)
{
    // Variable to store the final result
    long sum = 1;

    // Loop iterating through string
    for (int i = 0; i < s.size(); i++) {

        // If '$' is present at the even
        // position in the string
        if (i % 2 == 0 && s[i] == '$')

            //'sum' is multiplied by 21
            sum *= 21;

        // If '$' is present at the odd
        // position in the string
        else if (s[i] == '$')

            //'sum' is multiplied by 5
            sum *= 5;
    }

    return sum;
}

// Driver code
int main()
{
    // Let the string 'str' be s$$e$
    string str = "s$$e$";

    // Print result
    cout << countStrings(str) << endl;
    return 0;
}

Java

// Java implementation of above approach

import java.util.*;
import java.lang.*;
import java.io.*;

class GFG{

// Function to find the count of strings
static int countStrings(String s)
{
    // Variable to store the final result
    int sum = 1;

    // Loop iterating through string
    for (int i = 0; i < s.length(); i++) {

        // If '$' is present at the even
        // position in the string
        if (i % 2 == 0 && s.charAt(i) == '$')

            //'sum' is multiplied by 21
            sum *= 21;

        // If '$' is present at the odd
        // position in the string
        else if (s.charAt(i) == '$')

            //'sum' is multiplied by 5
            sum *= 5;
    }

    return sum;
}

// Driver code
public static void main(String args[])
{
    // Let the string 'str' be s$$e$
    String str = "s$$e$";

    // Print result
    System.out.println(countStrings(str));
}
}

Python 3

# Python 3 implementation of above approach

# Function to find the count of strings
def countStrings(s):

    # Variable to store the final result
    sum = 1

    # Loop iterating through string
    for i in range(len(s)):

        # If '$' is present at the even
        # position in the string
        if (i % 2 == 0 and s[i] == '$'):

            #'sum' is multiplied by 21
            sum *= 21

        # If '$' is present at the odd
        # position in the string
        elif(s[i] == '$'):

            # 'sum' is multiplied by 5
            sum *= 5

    return sum

# Driver code
if __name__ == "__main__":

    # Let the string 'str' be s$$e$
    str = "s$$e$"

    # Print result
    print(countStrings(str))

# this code is contributed by ChitraNayal

C#

// C# implementation of above approach

using System;

class GFG{

// Function to find the count of strings
static int countStrings(String s)
{
    // Variable to store the final result
    int sum = 1;

    // Loop iterating through string
    for (int i = 0; i < s.Length; i++) {

        // If '$' is present at the even
        // position in the string
        if (i % 2 == 0 && s[i] == '$')

            //'sum' is multiplied by 21
            sum *= 21;

        // If '$' is present at the odd
        // position in the string
        else if (s[i] == '$')

            //'sum' is multiplied by 5
            sum *= 5;
    }

    return sum;
}

// Driver code
public static void Main()
{
    // Let the string 'str' be s$$e$
    String str = "s$$e$";

    // Print result
    Console.WriteLine(countStrings(str));
}
}

PHP

<?php
// PHP implementation of above approach

// Function to find the count of strings
function countStrings($s)
{
    // Variable to store the
    // final result
    $sum = 1;

    // Loop iterating through string
    for ($i = 0; $i < strlen($s); $i++)
    {

        // If '$' is present at the even
        // position in the string
        if ($i % 2 == 0 && $s[$i] == '$')

            //'sum' is multiplied by 21
            $sum *= 21;

        // If '$' is present at the odd
        // position in the string
        else if ($s[$i] == '$')

            //'sum' is multiplied by 5
            $sum *= 5;
    }

    return $sum;
}

// Driver code

// Let the string 'str' be s$$e$
$str = "s\$\$e\$";

// Print result
echo countStrings($str);

// This code is contributed by Ryuga
?>

Javascript

<script>
// javascript implementation of above approach

// Function to find the count of strings
function countStrings( s)
{

    // Variable to store the final result
    let sum = 1;

    // Loop iterating through string
    for (let i = 0; i < s.length; i++)
    {

        // If '$' is present at the even
        // position in the string
        if (i % 2 == 0 && s[i] == '$')

            //'sum' is multiplied by 21
            sum *= 21;

        // If '$' is present at the odd
        // position in the string
        else if (s[i] == '$')

            //'sum' is multiplied by 5
            sum *= 5;
    }
    return sum;
}

// Driver code

    // Let the string 'str' be s$$e$
    let str = "s$$e$";

    // Print result
    document.write(countStrings(str));

// This code is contributed by gauravrajput1
</script>

Output

2205

Time complexity: O(n) where n is length of given string

Auxiliary space: O(1)