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: 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 ‘$’.
Finally, concatenate the three strings and print the newly concatenated string.
Below is the implementation of the above approach:
// CPP program to modify the given strings #include <iostream> #include <string.h> using namespace std;
// Function to modify the given three strings string modifyStr(string str1, string str2, string str3) { // Modifying first string
for ( int i = 0; i < str1.length(); i++) {
if (str1[i] == 'a' || str1[i] == 'e' ||
str1[i] == 'i' || str1[i] == 'o' ||
str1[i] == 'u' )
str1[i] = '*' ;
}
// Modifying third string
for ( int i = 0; i < str3.length(); i++) {
if (str3[i] != 'a' && str3[i] != 'e' &&
str3[i] != 'i' && str3[i] != 'o' &&
str3[i] != 'u' )
str3[i] = '$' ;
}
// 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 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)+ '$' +
str3.substring(i + 1 );
}
// Concatenating the three Strings
return (str1 + str2 + str3);
} // Driver code public static void main(String[] args)
{ String str1 = "how" ;
String str2 = "are" ;
String str3 = "you" ;
System.out.print(modifyStr(str1, str2, str3));
} } // This code is contributed by 29AjayKumar |
# Python3 program to modify the given Strings # Function to modify the given three Strings def modifyStr(str1, str2, str3):
# Modifying first String
for i in range ( len (str1)):
if (str1[i] = = 'a' or str1[i] = = 'e' or
str1[i] = = 'i' or str1[i] = = 'o'
or str1[i] = = 'u' ):
str1 = str1[ 0 :i] + '*' + str1[i + 1 :];
# Modifying third String
for i in range ( len (str3)):
if (str3[i] ! = 'a' and str3[i] ! = 'e' and str3[i] ! = 'i' and str3[i] ! = 'o'
and str3[i] ! = 'u' ):
str3 = str3[ 0 : i] + '$' + 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# 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)+ '$' +
str3.Substring(i + 1);
}
// Concatenating the three Strings
return (str1 + str2 + str3);
} // Driver code public static void Main(String[] args)
{ String str1 = "how" ;
String str2 = "are" ;
String str3 = "you" ;
Console.Write(modifyStr(str1, str2, str3));
} } // This code is contributed by PrinciRaj1992 |
<script> // Javascript program to modify the given Strings // Function to modify the given three Strings function modifyStr(str1, str2, str3)
{ // Modifying first String
for ( var i = 0; i < str1.length; i++)
{
if (str1.charAt(i) == 'a' ||
str1.charAt(i) == 'e' ||
str1.charAt(i) == 'i' ||
str1.charAt(i) == 'o' ||
str1.charAt(i) == 'u' )
str1 = str1.substring(0, i) + '*' +
str1.substring(i + 1);
}
// Modifying third String
for ( var i = 0; i < str3.length; i++)
{
if (str3.charAt(i) != 'a' &&
str3.charAt(i) != 'e' &&
str3.charAt(i) != 'i' &&
str3.charAt(i) != 'o' &&
str3.charAt(i) != 'u' )
str3 = str3.substring(0, i) + '$' +
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.