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 : neveropen 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.
C++
#include <iostream>#include <string>using namespace std;// Function to reverse a stringstring reverseString(string s) { string reversed; // Iterate over the original string in reverse order for (int i = s.length() - 1; i >= 0; i--) { // Append each character to a new string reversed += s[i]; } // Return the reversed string return reversed;}// Function to remove vowels from a stringstring removeVowels(string s,string reversed) { string withoutVowels; // Iterate over the string for (int i = 0; i < s.length(); i++) { char c = s[i]; // Check if the character is a vowel in original string if (c != 'a' && c != 'e' && c != 'i' && c != 'o' && c != 'u' && c != 'A' && c != 'E' && c != 'I' && c != 'O' && c != 'U') { // If not, append the reversed string character to a new string withoutVowels += reversed[i]; } } // Return the string without vowels return withoutVowels;}int main() { string s = "neveropen"; // Reverse the original string string reversed = reverseString(s); // Remove vowels from the reversed string string withoutVowels = removeVowels(s,reversed); // Print the resulting string without vowels cout << withoutVowels << endl; return 0;}// This code is contributed by Utkarsh. |
Java
public class Main { public static void main(String[] args) { String s = "neveropen"; // Reverse the original string String reversed = reverseString(s); // Remove vowels from the reversed string String withoutVowels = removeVowels(s, reversed); // Print the resulting string without vowels System.out.println(withoutVowels); } // Function to reverse a string public static String reverseString(String s) { String reversed = ""; // Iterate over the original string in reverse order for (int i = s.length() - 1; i >= 0; i--) { // Append each character to a new string reversed += s.charAt(i); } // Return the reversed string return reversed; } // Function to remove vowels from a string public static String removeVowels(String s, String reversed) { String withoutVowels = ""; // Iterate over the string for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); // Check if the character is a vowel in original string if (c != 'a' && c != 'e' && c != 'i' && c != 'o' && c != 'u' && c != 'A' && c != 'E' && c != 'I' && c != 'O' && c != 'U') { // If not, append the reversed string character to a new string withoutVowels += reversed.charAt(i); } } // Return the string without vowels return withoutVowels; }}//This Code is written by Sundaram. |
Python3
def reverseString(s): reversed_string = "" # Iterate over the original string in reverse order for i in range(len(s) - 1, -1, -1): # Append each character to a new string reversed_string += s[i] # Return the reversed string return reversed_stringdef removeVowels(s, reversed_string): withoutVowels = "" # Iterate over the string for i in range(len(s)): c = s[i] # Check if the character is a vowel in original string if c not in ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']: # If not, append the reversed string character to a new string withoutVowels += reversed_string[i] # Return the string without vowels return withoutVowelsif __name__ == "__main__": s = "neveropen" # Reverse the original string reversed_string = reverseString(s) # Remove vowels from the reversed string withoutVowels = removeVowels(s, reversed_string) # Print the resulting string without vowels print(withoutVowels) |
C#
using System;namespace ConsoleApp1 {class Program { // Function to reverse a string static string ReverseString(string s) { string reversed = ""; // Iterate over the original string in reverse order for (int i = s.Length - 1; i >= 0; i--) { // Append each character to a new string reversed += s[i]; } // Return the reversed string return reversed; } // Function to remove vowels from a string static string RemoveVowels(string s, string reversed) { string withoutVowels = ""; // Iterate over the string for (int i = 0; i < s.Length; i++) { char c = s[i]; // Check if the character is a vowel in original // string if (c != 'a' && c != 'e' && c != 'i' && c != 'o' && c != 'u' && c != 'A' && c != 'E' && c != 'I' && c != 'O' && c != 'U') { // If not, append the reversed string // character to a new string withoutVowels += reversed[i]; } } // Return the string without vowels return withoutVowels; } static void Main(string[] args) { string s = "neveropen"; // Reverse the original string string reversed = ReverseString(s); // Remove vowels from the reversed string string withoutVowels = RemoveVowels(s, reversed); // Print the resulting string without vowels Console.WriteLine(withoutVowels); }}}// This code is contributed by user_dtewbxkn77n |
Javascript
function reverseString(s) { let reversed = ""; // Iterate over the original string in reverse order for (let i = s.length - 1; i >= 0; i--) { // Append each character to a new string reversed += s[i]; } // Return the reversed string return reversed;}function removeVowels(s, reversed) { let withoutVowels = ""; // Iterate over the string for (let i = 0; i < s.length; i++) { let c = s[i]; // Check if the character is a vowel in original string if (c != 'a' && c != 'e' && c != 'i' && c != 'o' && c != 'u' && c != 'A' && c != 'E' && c != 'I' && c != 'O' && c != 'U') { // If not, append the reversed string character to a new string withoutVowels += reversed[i]; } } // Return the string without vowels return withoutVowels;}let s = "neveropen";// Reverse the original stringlet reversed = reverseString(s);// Remove vowels from the reversed stringlet withoutVowels = removeVowels(s, reversed);// Print the resulting string without vowelsconsole.log(withoutVowels); |
Output
segrfseg
Time complexity : O(n)
Auxiliary Space : O(n)
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 stringvoid 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 functionint main(){ string s = "neveropen"; int n = s.length(); replaceOriginal(s, n); return 0;} |
Java
// Java Program for removing characters// from reversed string where vowels are// present in original stringclass 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 = "neveropen"; 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 stringdef 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 Codeif __name__ == "__main__": s = "neveropen" 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 stringusing 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 = "neveropen"; 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 stringfunction 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 functionvar s = "neveropen";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)
Approach:
In this approach, we will iterate through each character of the string, and if the character is not a vowel, we will append it to a new string. Finally, we will reverse the new string and return it as the output.
Steps:
- Create an empty string called “new_string”.
- Iterate through each character of the input string.
- If the character is not a vowel, append it to “new_string”.
- Reverse the “new_string” using string slicing.
- Return the reversed string as the output.
Python3
def reverse_string_without_vowels(input_string): vowels = "AEIOUaeiou" new_string = "" for char in input_string: if char not in vowels: new_string += char return new_string[::-1]# Test the functioninput_string = "neveropen"output_string = reverse_string_without_vowels(input_string)print(output_string) |
Javascript
function reverseStringWithoutVowels(inputString) { const vowels = "AEIOUaeiou"; let newString = ""; for (let i = inputString.length - 1; i >= 0; i--) { if (!vowels.includes(inputString[i])) { newString += inputString[i]; } } return newString;}// Test the functionconst inputString = "neveropen";const outputString = reverseStringWithoutVowels(inputString);console.log(outputString); |
Output
skgrfskg
Time Complexity: The time complexity of this approach is O(n), where n is the length of the input string.
Auxiliary Space: The auxiliary space used in this approach is O(n), where n is the length of the input string.
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