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:

•    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;
}

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))

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));

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

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;
}

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;
}

Output

geks for geks

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

Auxiliary Space: O(1).