Print the given 3 string after modifying and concatenating

Given three strings(without spaces). The task is to print the new string after modifying the three given string as follows: 

• Replace all the vowels present in the first string with “*”.
• Don’t change anything in the second string.
• Replace all the consonants in the third string with “$”.
• Concatenate all of the three string to obtain the new string.

Examples: 

Input: how are you
Output\: h*ware$ou

Input: geeks for geeks
Output: g**ksfor$ee$$

Approach:
The idea is to traverse the first string and keep checking if any character is a vowel or not. Replace the character in the first string which is vowel with “*”. Similarly, traverse the third string and keep checking if any character is not a vowel. If a character in the third string is not a vowel(then it is a consonant), replace it with ‘$’.
Finally, concatenate the three strings and print the newly concatenated string. 

Below is the implementation of the above approach:

C++

// CPP program to modify the given strings
#include <iostream>
#include <string.h>
using namespace std;

// Function to modify the given three strings
string modifyStr(string str1, string str2, string str3)
{
    // Modifying first string
    for (int i = 0; i < str1.length(); i++) {
        if (str1[i] == 'a' || str1[i] == 'e' ||
            str1[i] == 'i' || str1[i] == 'o' ||
            str1[i] == 'u')
            str1[i] = '*';
    }

    // Modifying third string
    for (int i = 0; i < str3.length(); i++) {
        if (str3[i] != 'a' && str3[i] != 'e' &&
            str3[i] != 'i' && str3[i] != 'o' &&
            str3[i] != 'u')
            str3[i] = '$';
    }

    // Concatenating the three strings
    return (str1 + str2 + str3);
}

// Driver code
int main()
{
    string str1 = "how";
    string str2 = "are";
    string str3 = "you";

    cout << modifyStr(str1, str2, str3);

    return 0;
}

Java

// JAVA program to modify the given Strings
class GFG
{

// Function to modify the given three Strings
static String modifyStr(String str1, String str2, String str3)
{
    // Modifying first String
    for (int i = 0; i < str1.length(); i++) {
        if (str1.charAt(i) == 'a' || str1.charAt(i) == 'e' ||
            str1.charAt(i) == 'i' || str1.charAt(i) == 'o' ||
            str1.charAt(i) == 'u')
            str1 = str1.substring(0, i)+ '*' +
                   str1.substring(i + 1);
    }

    // Modifying third String
    for (int i = 0; i < str3.length(); i++) {
        if (str3.charAt(i) != 'a' && str3.charAt(i) != 'e' &&
            str3.charAt(i) != 'i' && str3.charAt(i) != 'o' &&
            str3.charAt(i) != 'u')
            str3 = str3.substring(0, i)+ '$' +
                   str3.substring(i + 1);
    }

    // Concatenating the three Strings
    return (str1 + str2 + str3);
}

// Driver code
public static void main(String[] args)
{
    String str1 = "how";
    String str2 = "are";
    String str3 = "you";

    System.out.print(modifyStr(str1, str2, str3));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to modify the given Strings

# Function to modify the given three Strings
def modifyStr(str1, str2, str3):

    # Modifying first String
    for i in range(len(str1)):
        if (str1[i] == 'a' or str1[i] == 'e' or
            str1[i] == 'i' or str1[i] == 'o'
            or str1[i] == 'u'):
            str1 = str1[0:i] + '*' + str1[i + 1:];


    # Modifying third String
    for i in range(len(str3)):
        if (str3[i] != 'a' and str3[i] != 'e' and
            str3[i] != 'i' and str3[i] != 'o'
            and str3[i] != 'u'):
            str3 = str3[0: i] + '$' + str3[i + 1:];

    # Concatenating the three Strings
    return (str1 + str2 + str3);

# Driver code
if __name__ == '__main__':
    str1 = "how";
    str2 = "are";
    str3 = "you";

    print(modifyStr(str1, str2, str3));

# This code is contributed by 29AjayKumar

C#

// C# program to modify the given Strings
using System;

class GFG
{

// Function to modify the given three Strings
static String modifyStr(String str1, String str2, String str3)
{
    // Modifying first String
    for (int i = 0; i < str1.Length; i++)
    {
        if (str1[i] == 'a' || str1[i] == 'e' ||
            str1[i] == 'i' || str1[i] == 'o' ||
            str1[i] == 'u')
            str1 = str1.Substring(0, i)+ '*' +
                str1.Substring(i + 1);
    }

    // Modifying third String
    for (int i = 0; i < str3.Length; i++)
    {
        if (str3[i] != 'a' && str3[i] != 'e' &&
            str3[i] != 'i' && str3[i] != 'o' &&
            str3[i] != 'u')
            str3 = str3.Substring(0, i)+ '$' +
                str3.Substring(i + 1);
    }

    // Concatenating the three Strings
    return (str1 + str2 + str3);
}

// Driver code
public static void Main(String[] args)
{
    String str1 = "how";
    String str2 = "are";
    String str3 = "you";

    Console.Write(modifyStr(str1, str2, str3));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program to modify the given Strings

// Function to modify the given three Strings
function modifyStr(str1, str2, str3)
{
    // Modifying first String
    for(var i = 0; i < str1.length; i++)
    {
        if (str1.charAt(i) == 'a' ||
            str1.charAt(i) == 'e' ||
            str1.charAt(i) == 'i' ||
            str1.charAt(i) == 'o' ||
            str1.charAt(i) == 'u')
            str1 = str1.substring(0, i) + '*' +
                   str1.substring(i + 1);
    }

    // Modifying third String
    for(var i = 0; i < str3.length; i++)
    {
        if (str3.charAt(i) != 'a' &&
            str3.charAt(i) != 'e' &&
            str3.charAt(i) != 'i' &&
            str3.charAt(i) != 'o' &&
            str3.charAt(i) != 'u')
            str3 = str3.substring(0, i) + '$' +
                   str3.substring(i + 1);
    }

    // Concatenating the three Strings
    return (str1 + str2 + str3);
}

// Driver code
var str1 = "how";
var str2 = "are";
var str3 = "you";

document.write(modifyStr(str1, str2, str3));

// This code is contributed by Ankita saini

</script>

Output:

h*ware$ou

Time Complexity: O(m+n), where m is the length of the first string and n is the length of the third string.

Auxiliary Space: O(1), as constant extra space is required.