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#

// 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)

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
?>

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>

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:

    Start the program by including the required header files and the standard namespace.

   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.

   Define the main function, which sets the input string to ” geeks for geeks”, calls the removeConsecutiveVowels function with this string, and outputs the result.

   End the program by returning 0 from the main function.

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;
}

C#

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.