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++
#include <bits/stdc++.h>
using namespace std;
bool is_vow(char c)
{
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
void removeVowels(string str)
{
printf("%c", str[0]);
for (int i = 1; str[i]; i++)
if ((!is_vow(str[i - 1])) ||
(!is_vow(str[i])))
printf("%c", str[i]);
}
int main()
{
char str[] = " geeks for geeks";
removeVowels(str);
}
|
Java
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG
{
static boolean is_vow(char c)
{
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
static void removeVowels(String str)
{
System.out.print(str.charAt(0));
for (int i = 1;
i < str.length(); i++)
if ((!is_vow(str.charAt(i - 1))) ||
(!is_vow(str.charAt(i))))
System.out.print(str.charAt(i));
}
public static void main(String[] args)
{
String str = "geeks for geeks";
removeVowels(str);
}
}
|
Python3
def is_vow(c):
return ((c == 'a') or (c == 'e') or
(c == 'i') or (c == 'o') or
(c == 'u'));
def removeVowels(str):
print(str[0], end = "");
for i in range(1,len(str)):
if ((is_vow(str[i - 1]) != True) or
(is_vow(str[i]) != True)):
print(str[i], end = "");
str= " geeks for geeks";
removeVowels(str);
|
C#
using System;
class GFG
{
static bool is_vow(char c)
{
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
static void removeVowels(string str)
{
Console.Write(str[0]);
for (int i = 1; i < str.Length; i++)
if ((!is_vow(str[i - 1])) ||
(!is_vow(str[i])))
Console.Write(str[i]);
}
static void Main()
{
string str = "geeks for geeks";
removeVowels(str);
}
}
|
Javascript
<script>
function is_vow(c)
{
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
function removeVowels(str)
{
document.write(str[0]);
for (let i = 1; i<str.length; i++)
if ((!is_vow(str[i - 1])) ||
(!is_vow(str[i])))
document.write(str[i]);
}
let str = " geeks for geeks";
removeVowels(str);
</script>
|
PHP
<?php
function is_vow($c)
{
return ($c == 'a') || ($c == 'e') ||
($c == 'i') || ($c == 'o') ||
($c == 'u');
}
function removeVowels($str)
{
printf($str[0]);
for ($i = 1; $i < strlen($str); $i++)
if ((!is_vow($str[$i - 1])) ||
(!is_vow($str[$i])))
printf($str[$i]);
}
$str= " geeks for geeks";
removeVowels($str);
?>
|
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) {
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 (isVowel(str[i])) {
if (!stk.empty() && isVowel(stk.top())) {
while (!stk.empty() && isVowel(stk.top())) {
stk.pop();
}
}
}
stk.push(str[i]);
}
string result = "";
while (!stk.empty()) {
result = stk.top() + result;
stk.pop();
}
return result;
}
int main() {
string str = " geeks for geeks";
cout << removeConsecutiveVowels(str) << endl;
return 0;
}
|
Java
import java.util.Stack;
public class RemoveConsecutiveVowels {
public static boolean isVowel(char c) {
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 (isVowel(str.charAt(i))) {
if (!stk.empty() && isVowel(stk.peek())) {
while (!stk.empty() && isVowel(stk.peek())) {
stk.pop();
}
}
}
stk.push(str.charAt(i));
}
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));
}
}
|
Python3
def is_vowel(c):
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 is_vowel(c):
if stack and is_vowel(stack[-1]):
while stack and is_vowel(stack[-1]):
stack.pop()
stack.append(c)
result = ""
while stack:
result = stack.pop() + result
return result
s = " geeks for geeks"
print(remove_consecutive_vowels(s))
|
C#
using System;
using System.Collections.Generic;
public class Program {
static bool IsVowel(char c)
{
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 (IsVowel(str[i])) {
if (stk.Count > 0 && IsVowel(stk.Peek())) {
while (stk.Count > 0
&& IsVowel(stk.Peek())) {
stk.Pop();
}
}
}
stk.Push(str[i]);
}
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));
}
}
|
Javascript
function isVowel(c) {
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 (isVowel(str[i]))
{
if (stk.length > 0 && isVowel(stk[stk.length - 1]))
{
while (stk.length > 0 && isVowel(stk[stk.length - 1])) {
stk.pop();
}
}
}
stk.push(str[i]);
}
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) {
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 (isVowel(str[i])) {
if (i == 0 || !isVowel(str[i - 1])) {
result += str[i];
}
} else {
result += str[i];
}
}
return result;
}
int main() {
string str = " geeks for geeks";
cout << removeConsecutiveVowels(str) << endl;
return 0;
}
|
Java
import java.io.*;
import java.util.HashSet;
public class GFG {
static boolean isVowel(char c) {
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 (isVowel(str.charAt(i))) {
if (i == 0 || !isVowel(str.charAt(i - 1))) {
result += str.charAt(i);
}
} else {
result += str.charAt(i);
}
}
return result;
}
public static void main(String[] args) {
String str = " geeks for geeks";
System.out.println(removeConsecutiveVowels(str));
}
}
|
Python3
def isVowel(c):
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 isVowel(string[i]):
if i == 0 or not isVowel(string[i - 1]):
result += string[i]
else:
result += string[i]
return result
string = ' geeks for geeks'
print(removeConsecutiveVowels(string))
|
Javascript
function isVowel(c) {
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 (isVowel(str[i])) {
if (i === 0 || !isVowel(str[i - 1])) {
result += str[i];
}
} else {
result += str[i];
}
}
return result;
}
const str = ' geeks for geeks';
console.log(removeConsecutiveVowels(str));
|
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) {
c = tolower(c);
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
void removeVowels(string str) {
string result = "";
result += str[0];
for (int i = 1; i < str.length(); i++) {
if ((!is_vowel(str[i - 1])) || (!is_vowel(str[i])))
result += str[i];
}
cout << result << endl;
}
int main() {
string str = "geeks for geeks";
removeVowels(str);
return 0;
}
|
Time Complexity: O(n), where n is the length of the input string.
Auxiliary Space: O(1).