Print reverse string after removing vowels
Given a string s, print reverse of string and remove the characters from the reversed string where there are vowels in the original string.
Examples:
Input : geeksforgeeks Output : segrfseg Explanation : Reversed string is skeegrofskeeg, removing characters from indexes 1, 2, 6, 9 & 10 (0 based indexing), we get segrfseg . Input :duck Output :kud
A simple solution is to first reverse the string, then traverse the reversed string and remove vowels.
An efficient solution is to do both tasks in one traversal.
Create an empty string r and traverse the original string s and assign the value to the string r. Check whether, at that index, the original string contains a consonant or not. If yes then print the element at that index from string r.
Basic implementation of the above approach :
C++
// CPP Program for removing characters // from reversed string where vowels are // present in original string #include <bits/stdc++.h> using namespace std; // Function for replacing the string void replaceOriginal(string s, int n) { // initialize a string of length n string r(n, ' ' ); // Traverse through all characters of string for ( int i = 0; i < n; i++) { // assign the value to string r // from last index of string s r[i] = s[n - 1 - i]; // if s[i] is a consonant then // print r[i] if (s[i] != 'a' && s[i] != 'e' && s[i] != 'i' && s[i] != 'o' && s[i] != 'u' ) { cout << r[i]; } } cout << endl; } // Driver function int main() { string s = "geeksforgeeks" ; int n = s.length(); replaceOriginal(s, n); return 0; } |
Java
// Java Program for removing characters // from reversed string where vowels are // present in original string class GFG { // Function for replacing the string static void replaceOriginal(String s, int n) { // initialize a string of length n char r[] = new char [n]; // Traverse through all characters of string for ( int i = 0 ; i < n; i++) { // assign the value to string r // from last index of string s r[i] = s.charAt(n - 1 - i); // if s[i] is a consonant then // print r[i] if (s.charAt(i) != 'a' && s.charAt(i) != 'e' && s.charAt(i) != 'i' && s.charAt(i) != 'o' && s.charAt(i) != 'u' ) { System.out.print(r[i]); } } System.out.println( "" ); } // Driver function public static void main(String[] args) { String s = "geeksforgeeks" ; int n = s.length(); replaceOriginal(s, n); } } // This code is contributed by princiRaj1992 |
Python3
# Python3 Program for removing characters # from reversed string where vowels are # present in original string # Function for replacing the string def replaceOriginal(s, n): # initialize a string of length n r = [ ' ' ] * n # Traverse through all characters of string for i in range (n): # assign the value to string r # from last index of string s r[i] = s[n - 1 - i] # if s[i] is a consonant then # print r[i] if (s[i] ! = 'a' and s[i] ! = 'e' and s[i] ! = 'i' and s[i] ! = 'o' and s[i] ! = 'u' ): print (r[i], end = "") print () # Driver Code if __name__ = = "__main__" : s = "geeksforgeeks" n = len (s) replaceOriginal(s, n) # This code is contributed by # sanjeev2552 |
C#
// C# Program for removing characters // from reversed string where vowels are // present in original string using System; class GFG { // Function for replacing the string static void replaceOriginal(String s, int n) { // initialize a string of length n char []r = new char [n]; // Traverse through all characters of string for ( int i = 0; i < n; i++) { // assign the value to string r // from last index of string s r[i] = s[n - 1 - i]; // if s[i] is a consonant then // print r[i] if (s[i] != 'a' && s[i] != 'e' && s[i] != 'i' && s[i] != 'o' && s[i] != 'u' ) { Console.Write(r[i]); } } Console.WriteLine( "" ); } // Driver code public static void Main(String[] args) { String s = "geeksforgeeks" ; int n = s.Length; replaceOriginal(s, n); } } // This code is contributed by Rajput-JI |
Javascript
<script> // JavaScript Program for removing characters // from reversed string where vowels are // present in original string // Function for replacing the string function replaceOriginal(s, n) { // initialize a string of length n var r = new Array(n); // Traverse through all characters of string for ( var i = 0; i < n; i++) { // assign the value to string r // from last index of string s r[i] = s.charAt(n - 1 - i); // if s[i] is a consonant then // print r[i] if (s.charAt(i) != 'a' && s.charAt(i) != 'e' && s.charAt(i) != 'i' && s.charAt(i) != 'o' && s.charAt(i) != 'u' ) { document.write(r[i]); } } document.write( "" ); } // Driver function var s = "geeksforgeeks" ; var n = s.length; replaceOriginal(s, n); // This code is contributed by shivanisinghss2110 </script> |
Output
segrfseg
Complexity Analysis:
- Time complexity : O(n)
- Auxiliary Space : O(n)
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