Remove consecutive vowels from string

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

// 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

# 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)

Javascript

<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
// 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 
?>

Output

geks for geks

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;
}

Java

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

Python3

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

Javascript

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));

Output

 geks for geks

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;
}

Java

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)

Python3

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"

    }
}

Javascript

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

// 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

Python3

# 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);

    }
}

Javascript

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);

Output

geks for geks

Time Complexity: O(n), where n is the length of the input string.
Auxiliary Space: O(1).

Article Tags :

DSA

Misc

Strings

Basic Coding Problems

vowel-consonant