Remove consecutive vowels from string
Last Updated :
30 Nov, 2023
Given a string s of lowercase letters, we need to remove consecutive vowels from the string
Note : Sentence should not contain two consecutive vowels ( a, e, i, o, u).
Examples :
Input: geeks for geeks
Output: geks for geks
Input : your article is in queue
Output : yor article is in qu
Approach: Iterate string using a loop and check for the repetitiveness of vowels in a given sentence and in case if consecutive vowels are found then delete the vowel till coming next consonant and printing the updated string.
Implementation:
C++
// C++ program for printing sentence
// without repetitive vowels
#include <bits/stdc++.h>
using namespace std;
// function which returns True or False
// for occurrence of a vowel
bool is_vow(char c)
{
// this compares vowel with
// character 'c'
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
// function to print resultant string
void removeVowels(string str)
{
// print 1st character
printf("%c", str[0]);
// loop to check for each character
for (int i = 1; str[i]; i++)
// comparison of consecutive characters
if ((!is_vow(str[i - 1])) ||
(!is_vow(str[i])))
printf("%c", str[i]);
}
// Driver Code
int main()
{
char str[] = " geeks for geeks";
removeVowels(str);
}
// This code is contributed by Abhinav96
// Java program for printing sentence
// without repetitive vowels
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG
{
// function which returns
// True or False for
// occurrence of a vowel
static boolean is_vow(char c)
{
// this compares vowel
// with character 'c'
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
// function to print
// resultant string
static void removeVowels(String str)
{
// print 1st character
System.out.print(str.charAt(0));
// loop to check for
// each character
for (int i = 1;
i < str.length(); i++)
// comparison of
// consecutive characters
if ((!is_vow(str.charAt(i - 1))) ||
(!is_vow(str.charAt(i))))
System.out.print(str.charAt(i));
}
// Driver Code
public static void main(String[] args)
{
String str = "geeks for geeks";
removeVowels(str);
}
}
# Python3 implementation for printing
# sentence without repetitive vowels
# function which returns True or False
# for occurrence of a vowel
def is_vow(c):
# this compares vowel with
# character 'c'
return ((c == 'a') or (c == 'e') or
(c == 'i') or (c == 'o') or
(c == 'u'));
# function to print resultant string
def removeVowels(str):
# print 1st character
print(str[0], end = "");
# loop to check for each character
for i in range(1,len(str)):
# comparison of consecutive
# characters
if ((is_vow(str[i - 1]) != True) or
(is_vow(str[i]) != True)):
print(str[i], end = "");
# Driver code
str= " geeks for geeks";
removeVowels(str);
# This code is contributed by mits
// C# program for printing sentence
// without repetitive vowels
using System;
class GFG
{
// function which returns
// True or False for
// occurrence of a vowel
static bool is_vow(char c)
{
// this compares vowel
// with character 'c'
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
// function to print
// resultant string
static void removeVowels(string str)
{
// print 1st character
Console.Write(str[0]);
// loop to check for
// each character
for (int i = 1; i < str.Length; i++)
// comparison of
// consecutive characters
if ((!is_vow(str[i - 1])) ||
(!is_vow(str[i])))
Console.Write(str[i]);
}
// Driver Code
static void Main()
{
string str = "geeks for geeks";
removeVowels(str);
}
}
// This code is contributed
// by Manish Shaw(manishshaw1)
<script>
// JavaScript program for printing sentence
// without repetitive vowels
// function which returns True or False
// for occurrence of a vowel
function is_vow(c)
{
// this compares vowel with
// character 'c'
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
// function to print resultant string
function removeVowels(str)
{
// print 1st character
document.write(str[0]);
// loop to check for each character
for (let i = 1; i<str.length; i++)
// comparison of consecutive characters
if ((!is_vow(str[i - 1])) ||
(!is_vow(str[i])))
document.write(str[i]);
}
// Driver Code
let str = " geeks for geeks";
removeVowels(str);
// This code is contributed by shinjanpatra
</script>
<?php
// PHP implementation for printing
// sentence without repetitive vowels
// function which returns True or False
// for occurrence of a vowel
function is_vow($c)
{
// this compares vowel with
// character 'c'
return ($c == 'a') || ($c == 'e') ||
($c == 'i') || ($c == 'o') ||
($c == 'u');
}
// function to print resultant string
function removeVowels($str)
{
// print 1st character
printf($str[0]);
// loop to check for each character
for ($i = 1; $i < strlen($str); $i++)
// comparison of consecutive
// characters
if ((!is_vow($str[$i - 1])) ||
(!is_vow($str[$i])))
printf($str[$i]);
}
// Driver code
$str= " geeks for geeks";
removeVowels($str);
// This code is contributed by mits
?>
Time Complexity: O(n), where n is the length of the string
Space Complexity: O(n), where n is the length of the string
Another approach:- here's another approach in C++ to remove consecutive vowels from a string using a stack:
- Define a function named isVowel that takes a character as input and returns a boolean value indicating whether the character is a vowel.
- Define a function named removeConsecutiveVowels that takes a string as input and returns a string with all consecutive vowels removed.
- Create a stack named stk to store the characters of the input string.
- Get the length of the input string.
- Loop through each character of the input string by using a for loop.
- Check if the current character is a vowel by calling the isVowel function.
- If the current character is a vowel, check if the stack is not empty and the top of the stack is also a vowel.
- If the conditions in step 8 are satisfied, pop all consecutive vowels from the stack.
- Push the current character onto the stack.
- Construct the result string by popping all elements from the stack.
- Return the result string.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <string>
#include <stack>
using namespace std;
bool isVowel(char c) {
// check if a character is a vowel
return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
}
string removeConsecutiveVowels(string str) {
stack<char> stk;
int len = str.length();
for (int i = 0; i < len; i++) {
// if current character is a vowel
if (isVowel(str[i])) {
// check if the stack is not empty and the top of the stack is also a vowel
if (!stk.empty() && isVowel(stk.top())) {
// pop all consecutive vowels from the stack
while (!stk.empty() && isVowel(stk.top())) {
stk.pop();
}
}
}
// push the current character onto the stack
stk.push(str[i]);
}
// construct the result string by popping all elements from the stack
string result = "";
while (!stk.empty()) {
result = stk.top() + result;
stk.pop();
}
return result;
}
int main() {
string str = " geeks for geeks";
cout << removeConsecutiveVowels(str) << endl; // expected output: "ltcdsccmmntyfrcdrs"
return 0;
}
import java.util.Stack;
public class RemoveConsecutiveVowels {
public static boolean isVowel(char c) {
// check if a character is a vowel
return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
}
public static String removeConsecutiveVowels(String str) {
Stack<Character> stk = new Stack<>();
int len = str.length();
for (int i = 0; i < len; i++) {
// if current character is a vowel
if (isVowel(str.charAt(i))) {
// check if the stack is not empty and the top of the stack is also a vowel
if (!stk.empty() && isVowel(stk.peek())) {
// pop all consecutive vowels from the stack
while (!stk.empty() && isVowel(stk.peek())) {
stk.pop();
}
}
}
// push the current character onto the stack
stk.push(str.charAt(i));
}
// construct the result string by popping all elements from the stack
StringBuilder result = new StringBuilder();
while (!stk.empty()) {
result.insert(0, stk.peek());
stk.pop();
}
return result.toString();
}
public static void main(String[] args) {
String str = " geeks for geeks";
System.out.println(removeConsecutiveVowels(str)); // expected output: "ltcdsccmmntyfrcdrs"
}
}
// This code is contributed by Prajwal Kandekar
def is_vowel(c):
# check if a character is a vowel
return (c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u' or
c == 'A' or c == 'E' or c == 'I' or c == 'O' or c == 'U')
def remove_consecutive_vowels(s):
stack = []
for c in s:
# if current character is a vowel
if is_vowel(c):
# check if the stack is not empty and the top of the stack is also a vowel
if stack and is_vowel(stack[-1]):
# pop all consecutive vowels from the stack
while stack and is_vowel(stack[-1]):
stack.pop()
# push the current character onto the stack
stack.append(c)
# construct the result string by popping all elements from the stack
result = ""
while stack:
result = stack.pop() + result
return result
# test the function
s = " geeks for geeks"
print(remove_consecutive_vowels(s))
using System;
using System.Collections.Generic;
public class Program {
static bool IsVowel(char c)
{
// check if a character is a vowel
return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
|| c == 'u' || c == 'A' || c == 'E'
|| c == 'I' || c == 'O' || c == 'U');
}
static string RemoveConsecutiveVowels(string str)
{
Stack<char> stk = new Stack<char>();
int len = str.Length;
for (int i = 0; i < len; i++) {
// if current character is a vowel
if (IsVowel(str[i])) {
// check if the stack is not empty and the
// top of the stack is also a vowel
if (stk.Count > 0 && IsVowel(stk.Peek())) {
// pop all consecutive vowels from the
// stack
while (stk.Count > 0
&& IsVowel(stk.Peek())) {
stk.Pop();
}
}
}
// push the current character onto the stack
stk.Push(str[i]);
}
// construct the result string by popping all
// elements from the stack
string result = "";
while (stk.Count > 0) {
result = stk.Peek() + result;
stk.Pop();
}
return result;
}
public static void Main()
{
string str = " geeks for geeks";
Console.WriteLine(RemoveConsecutiveVowels(
str)); // expected output: " gks fr gks"
}
}
// This code is contributed by user_dtewbxkn77n
function isVowel(c) {
// check if a character is a vowel
return (c === 'a' || c === 'e' || c === 'i' || c === 'o' || c === 'u' ||
c === 'A' || c === 'E' || c === 'I' || c === 'O' || c === 'U');
}
function removeConsecutiveVowels(str) {
let stk = [];
let len = str.length;
for (let i = 0; i < len; i++)
{
// if current character is a vowel
if (isVowel(str[i]))
{
// check if the stack is not empty and the top of the stack is also a vowel
if (stk.length > 0 && isVowel(stk[stk.length - 1]))
{
// pop all consecutive vowels from the stack
while (stk.length > 0 && isVowel(stk[stk.length - 1])) {
stk.pop();
}
}
}
// push the current character onto the stack
stk.push(str[i]);
}
// construct the result string by popping all elements from the stack
let result = "";
while (stk.length > 0) {
result = stk[stk.length - 1] + result;
stk.pop();
}
return result;
}
let str = " geeks for geeks";
console.log(removeConsecutiveVowels(str));
Time Complexity: O(n), where n is the length of the string
The time complexity of the removeConsecutiveVowels function is O(n), where n is the length of the input string. This is because each character of the input string is processed once in the for loop, and all operations inside the loop are constant time operations.
Space Complexity: O(n), where n is the length of the string
The space complexity of the function is O(n), where n is the length of the input string. This is because the size of the stack can be at most the length of the input string, and the result string can also be of the same size as the input string in the worst case.
Another Approach:
This approach works by iterating over the input string and checking each character. If the current character is a vowel, it checks whether the previous character is also a vowel. If the previous character is not a vowel, it appends the current character to the result string. If the previous character is a vowel, it skips over the current character and continues iterating. If the current character is not a vowel, it simply appends it to the result string.
This approach does not use a stack, which can make it simpler and easier to understand. However, it may be slightly less efficient than the stack-based approach, since it needs to check the previous character at each iteration.
C++
#include <iostream>
#include <string>
using namespace std;
bool isVowel(char c) {
// check if a character is a vowel
return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
}
string removeConsecutiveVowels(string str) {
string result = "";
int len = str.length();
for (int i = 0; i < len; i++) {
// if current character is a vowel
if (isVowel(str[i])) {
// check if the previous character is also a vowel
if (i == 0 || !isVowel(str[i - 1])) {
// if not, append the current character to the result string
result += str[i];
}
} else {
// if the current character is not a vowel, append it to the result string
result += str[i];
}
}
return result;
}
int main() {
string str = " geeks for geeks";
cout << removeConsecutiveVowels(str) << endl; // expected output: " ltcdsccmmntyfrcdrs"
return 0;
}
import java.io.*;
import java.util.HashSet;
public class GFG {
static boolean isVowel(char c) {
// Check if a character is a vowel
return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');
}
static String removeConsecutiveVowels(String str) {
String result = "";
int len = str.length();
for (int i = 0; i < len; i++) {
// If current character is a vowel
if (isVowel(str.charAt(i))) {
// Check if the previous character is also a vowel
if (i == 0 || !isVowel(str.charAt(i - 1))) {
// If not, append the current character to the result string
result += str.charAt(i);
}
} else {
// If the current character is not a vowel, append it to the result string
result += str.charAt(i);
}
}
return result;
}
public static void main(String[] args) {
String str = " geeks for geeks";
System.out.println(removeConsecutiveVowels(str)); // expected output: " ltcdsccmmntyfrcdrs"
}
}
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
def isVowel(c):
# Check if a character is a vowel
return c in ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
def removeConsecutiveVowels(string):
result = ''
length = len(string)
for i in range(length):
# If current character is a vowel
if isVowel(string[i]):
# Check if the previous character is also a vowel
if i == 0 or not isVowel(string[i - 1]):
# If not, append the current character to
# the result string
result += string[i]
else:
# If the current character is not a vowel,
# append it to the result string
result += string[i]
return result
# Driver code
string = ' geeks for geeks'
print(removeConsecutiveVowels(string))
# THIS CODE IS CONTRIBUTED BY KANCHAN AGARWAL
using System;
class Program {
static bool IsVowel(char c)
{
// Check if a character is a vowel
return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
|| c == 'u' || c == 'A' || c == 'E'
|| c == 'I' || c == 'O' || c == 'U');
}
static string RemoveConsecutiveVowels(string str)
{
string result = "";
int len = str.Length;
for (int i = 0; i < len; i++) {
// If the current character is a vowel
if (IsVowel(str[i])) {
// Check if the previous character is also a
// vowel
if (i == 0 || !IsVowel(str[i - 1])) {
// If not, append the current character
// to the result string
result += str[i];
}
}
else {
// If the current character is not a vowel,
// append it to the result string
result += str[i];
}
}
return result;
}
static void Main(string[] args)
{
string str = " geeks for geeks";
Console.WriteLine(RemoveConsecutiveVowels(
str)); // Expected output: " ltcdsccmmntyfrcdrs"
}
}
function isVowel(c) {
// Check if a character is a vowel
return (
c === 'a' ||
c === 'e' ||
c === 'i' ||
c === 'o' ||
c === 'u' ||
c === 'A' ||
c === 'E' ||
c === 'I' ||
c === 'O' ||
c === 'U'
);
}
function removeConsecutiveVowels(str) {
let result = '';
const len = str.length;
for (let i = 0; i < len; i++) {
// If current character is a vowel
if (isVowel(str[i])) {
// Check if the previous character is also a vowel
if (i === 0 || !isVowel(str[i - 1])) {
// If not, append the current character to the
// result string
result += str[i];
}
} else {
// If the current character is not a vowel, append it
// to the result string
result += str[i];
}
}
return result;
}
// Driver code
const str = ' geeks for geeks';
console.log(removeConsecutiveVowels(str));
// Expected output: " ltcdsccmmntyfrcdrs"
Output:
geks for geks
Time Complexity: O(n), where n is the length of the input string.
Space Complexity: O(n), where n is the length of the input string.
Constant Space Approach:
The algorithm of the constant space approach for removing repetitive vowels from a sentence is as follows:
- Initialize an empty string called result to store the modified string without repetitive vowels.
- Add the first character of the input string str to the result string.
- Iterate through the remaining characters of the input string from the second character onward.
- For each character at index i in the input string:
- Check if the current character str[i] and the previous character str[i-1] are both vowels using the is_vowel function.
If both characters are vowels, skip adding the current character to the result string, as it is a repetitive vowel.
If either the current character or the previous character is not a vowel, add the current character to the result string.
After iterating through all the characters in the input string, the result string will contain the modified string without repetitive vowels. - Print the result string as the output.
Here is the code of above approach:
C++
#include <iostream>
using namespace std;
bool is_vowel(char c) {
// Convert character to lowercase for case-insensitive comparison
c = tolower(c);
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
void removeVowels(string str) {
// Initialize the result string with the first character of the input string
string result = "";
result += str[0];
// Loop to check for each character starting from the second character
for (int i = 1; i < str.length(); i++) {
// Comparison of consecutive characters
if ((!is_vowel(str[i - 1])) || (!is_vowel(str[i])))
result += str[i];
}
// Print the resultant string
cout << result << endl;
}
int main() {
string str = "geeks for geeks";
removeVowels(str);
return 0;
}
// Java code for the above approach
public class GFG {
// Function to check if a character is a vowel
public static boolean isVowel(char c)
{
// Convert character to lowercase for
// case-insensitive comparison
c = Character.toLowerCase(c);
return (c == 'a') || (c == 'e') || (c == 'i')
|| (c == 'o') || (c == 'u');
}
// Function to remove vowels from a string
public static void removeVowels(String str)
{
// Initialize the result string with the first
// character of the input string
StringBuilder result = new StringBuilder();
result.append(str.charAt(0));
// Loop to check for each character starting from
// the second character
for (int i = 1; i < str.length(); i++) {
// Comparison of consecutive characters
if ((!isVowel(str.charAt(i - 1)))
|| (!isVowel(str.charAt(i)))) {
result.append(str.charAt(i));
}
}
// Print the resultant string
System.out.println(result);
}
// Driver code
public static void main(String[] args)
{
String str = "geeks for geeks";
removeVowels(str);
}
}
// This code is contributed by Susobhan Akhuli
# Python code for the above approach
def is_vowel(c):
# Convert character to lowercase for case-insensitive comparison
c = c.lower()
return c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u'
def removeVowels(string):
# Initialize the result string with the first character of the input string
result = string[0]
# Loop to check for each character starting from the second character
for i in range(1, len(string)):
# Comparison of consecutive characters
if (not is_vowel(string[i - 1])) or (not is_vowel(string[i])):
result += string[i]
# Print the resultant string
print(result)
if __name__ == "__main__":
string = "geeks for geeks"
removeVowels(string)
# This code is contributed by Susobhan Akhuli
using System;
class Program
{
// Function to check if a character is a vowel
static bool IsVowel(char c)
{
// Convert character to lowercase for case-insensitive comparison
c = char.ToLower(c);
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
// Function to remove vowels from a string
static void RemoveVowels(string str)
{
// Initialize the result string with the first character of the input string
string result = "" + str[0];
// Loop to check for each character starting from the second character
for (int i = 1; i < str.Length; i++)
{
// Comparison of consecutive characters
if ((!IsVowel(str[i - 1])) || (!IsVowel(str[i])))
result += str[i];
}
// Print the resultant string
Console.WriteLine(result);
}
static void Main()
{
string str = "geeks for geeks";
RemoveVowels(str);
}
}
function isVowel(c) {
// Convert character to lowercase for case-insensitive comparison
c = c.toLowerCase();
return (c === 'a') || (c === 'e') ||
(c === 'i') || (c === 'o') ||
(c === 'u');
}
function removeVowels(str) {
// Initialize the result string with the first character of the input string
let result = str[0];
// Loop to check for each character starting from the second character
for (let i = 1; i < str.length; i++) {
// Comparison of consecutive characters
if ((!isVowel(str[i - 1])) || (!isVowel(str[i]))) {
result += str[i];
}
}
// Print the resultant string
console.log(result);
}
// Driver code
let str = "geeks for geeks";
removeVowels(str);
Time Complexity: O(n), where n is the length of the input string.
Auxiliary Space: O(1).
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Check if the given string is vowel prime
Given a string str of lowercase English alphabets, the task is to check whether the string is a vowel prime or not. A string is said to be vowel prime if all the vowels in the string appear at only prime indices.Examples: Input: str = "geeksforgeeks" Output: No str[1] = 'e' is a vowel but 1 is not p
8 min read
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 : geeksforgeeksOutput : segrfsegExplanation :Reversed string is skeegrofskeeg, removing characters from indexes 1, 2, 6, 9 & 10 (0 based ind
13 min read
Count of substrings consisting only of vowels
Given a string S, the task is to count all substrings which contain only vowels. Examples: Input: S = "geeksforgeeks" Output: 7 Explanation: Substrings {"e", "ee", "e", "o", "e", "ee", "e"} consists only of vowels. Input: S = "aecui" Output: 6 Explanation: Substrings {"a", "ae", "e", "u", "ui", "i"}
11 min read
Program to remove vowels from Linked List
Given a singly linked list, the task is to remove the vowels from the given linked list. Examples: Input: g -> e -> e -> k -> s -> f -> o -> r -> g -> e -> e -> k -> s Output: g -> k -> s -> f -> r -> g -> k -> s Explanation: After removing vo
15 min read
Longest substring of vowels
Given a string s of lowercase letters, we need to find the longest substring length that contains (a, e, i, o, u) only. Examples : Input: s = "geeksforgeeks" Output: 2 Longest substring is "ee" Input: s = "theeare" Output: 3 The idea is to traverse the string and keep track of the current number of
5 min read
Count the pairs of vowels in the given string
Given a string str consisting of lowercase English alphabets, the task is to count the number of adjacent pairs of vowels.Examples: Input: str = "abaebio" Output: 2 (a, e) and (i, o) are the only valid pairs.Input: str = "aeoui" Output: 4 Approach: Starting from the first character of the string to
5 min read
Count of substrings consisting of even number of vowels
Given a string S of length N, the task is to find the number of non-empty substrings having even number of vowels. Examples: Input: N = 5, S = "abcde"Output: 7Explanation: All possible substrings with even number of vowels are:Substring Vowels{abcde} 2{b} 0{bc} 0{bcd} 0{c} 0{cd} 0{d} 0Input: N=4, S=
11 min read
Remove even frequency characters from the string
Given a string 'str', the task is to remove all the characters from the string that have even frequencies. Examples: Input: str = "aabbbddeeecc" Output: bbbeee The characters a, d, c have even frequencies So, they are removed from the string. Input: str = "zzzxxweeerr" Output: zzzweee Approach: Crea
8 min read