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 “$”.
- Concatenate all of the three string to obtain the new string.
Examples:
Input: how are you
Output\: h*ware$ouInput: zambiatek for zambiatek
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> |
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.
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