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.
#include <iostream>
#include <string>
using namespace std;
// Function to reverse a string
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
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;
}
int main() {
string s = "geeksforgeeks";
// 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.
public class Main {
public static void main(String[] args) {
String s = "geeksforgeeks";
// 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.
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_string
def 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 withoutVowels
if __name__ == "__main__":
s = "geeksforgeeks"
# 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)
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 = "geeksforgeeks";
// 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
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 = "geeksforgeeks";
// Reverse the original string
let reversed = reverseString(s);
// Remove vowels from the reversed string
let withoutVowels = removeVowels(s, reversed);
// Print the resulting string without vowels
console.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 :
// 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 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 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# 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
<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)
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.
#include <iostream>
#include <string>
#include<bits/stdc++.h>
// Function to reverse a string without vowels
std::string reverseStringWithoutVowels(const std::string& inputString) {
// Define vowels
std::string vowels = "AEIOUaeiou";
// Initialize an empty string to store the result
std::string newString = "";
// Iterate through each character in the input string
for (char ch : inputString) {
// Check if the character is not a vowel
if (vowels.find(ch) == std::string::npos) {
// Append the character to the new string
newString += ch;
}
}
// Reverse the new string
std::reverse(newString.begin(), newString.end());
// Return the reversed string without vowels
return newString;
}
int main() {
// Test the function
std::string inputString = "geeksforgeeks";
std::string outputString = reverseStringWithoutVowels(inputString);
std::cout << outputString << std::endl;
return 0;
}
// This code is contributed by shivamgupta0987654321
import java.util.Scanner;
public class ReverseStringWithoutVowels {
// Function to reverse a string without vowels
static String reverseStringWithoutVowels(String inputString) {
// Define vowels
String vowels = "AEIOUaeiou";
// Initialize a StringBuilder to store the result
StringBuilder newString = new StringBuilder();
// Iterate through each character in the input string
for (char ch : inputString.toCharArray()) {
// Check if the character is not a vowel
if (vowels.indexOf(ch) == -1) {
// Append the character to the StringBuilder
newString.append(ch);
}
}
// Reverse the StringBuilder
newString.reverse();
// Return the reversed string without vowels
return newString.toString();
}
public static void main(String[] args) {
// Test the function
String inputString = "geeksforgeeks";
String outputString = reverseStringWithoutVowels(inputString);
System.out.println(outputString);
}
}
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 function
input_string = "geeksforgeeks"
output_string = reverse_string_without_vowels(input_string)
print(output_string)
using System;
class Program
{
// Function to reverse a string after removing vowels
static string ReverseStringWithoutVowels(string inputString)
{
// Define a string containing all the vowels in uppercase and lowercase
string vowels = "AEIOUaeiou";
// Initialize a string to store the characters without vowels
string newString = "";
// Iterate through each character in the input string
foreach (char c in inputString)
{
// Check if the character is not a vowel
if (vowels.IndexOf(c) == -1)
{
// If it's not a vowel, add it to the new string
newString += c;
}
}
// Convert the new string to a character array
char[] charArray = newString.ToCharArray();
// Reverse the character array
Array.Reverse(charArray);
// Convert the reversed character array back to a string
return new string(charArray);
}
static void Main()
{
// Test the ReverseStringWithoutVowels function
string inputString = "geeksforgeeks";
string outputString = ReverseStringWithoutVowels(inputString);
// Print the reversed string without vowels
Console.WriteLine(outputString);
}
}
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 function
const inputString = "geeksforgeeks";
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.