Count the number of vowels and consonants in a Linked List
Last Updated :
11 Feb, 2022
Given a linked list containing lowercase English alphabets, the task is to count the number of consonants and vowels present in the linked list.
Example:
Input: Linked List: a ->b->o->y -> e ->z->NULL
Output:
Vowels : 3
Consonants: 3
Input: Linked List: a -> e -> b->c->s->e->y->t->NULL
Output:
Vowels: 3
Consonants:5
Approach: To solve this problem, follow the below steps:
- Create two variables, vowel and consonant to store the number of vowels and consonants respectively. Initialize both of them with 0.
- Now, start traversing the linked list, and increment vowel by 1 if the character matches with anyone from the set of [a, e, i, o, u] else increment consonant by 1.
- Print the answer according to the above observation.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
char data;
struct Node* next;
Node(char key)
{
data = key;
next = NULL;
}
};
bool isVowel(char x)
{
return (x == 'a' || x == 'e'
|| x == 'i' || x == 'o'
|| x == 'u');
}
void count(struct Node* head)
{
int vowel = 0;
int consonant = 0;
for (Node* itr = head; itr != NULL; itr = itr->next) {
if (isVowel(itr->data)) {
vowel++;
}
else {
consonant++;
}
}
cout << "Vowel: " << vowel << endl;
cout << "Consonant: " << consonant << endl;
}
int main()
{
Node* head = new Node('u');
head->next = new Node('h');
head->next->next = new Node('d');
head->next->next->next = new Node('a');
head->next->next->next->next = new Node('n');
count(head);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static class Node {
char data;
Node next;
Node(char key)
{
data = key;
next = null;
}
};
static boolean isVowel(char x)
{
return (x == 'a' || x == 'e'
|| x == 'i' || x == 'o'
|| x == 'u');
}
static void count(Node head)
{
int vowel = 0;
int consonant = 0;
for (Node itr = head; itr != null; itr = itr.next) {
if (isVowel(itr.data)) {
vowel++;
}
else {
consonant++;
}
}
System.out.print("Vowel: " + vowel +"\n");
System.out.print("Consonant: " + consonant +"\n");
}
public static void main(String[] args)
{
Node head = new Node('u');
head.next = new Node('h');
head.next.next = new Node('d');
head.next.next.next = new Node('a');
head.next.next.next.next = new Node('n');
count(head);
}
}
|
Python3
class Node:
def __init__(self, data):
self.data = data;
self.next = None;
def isVowel(x):
return (x == 'a' or x == 'e' or x == 'i' or x == 'o' or x == 'u');
def count(head):
vowel = 0;
consonant = 0;
while(head!=None):
if (isVowel(head.data)):
vowel += 1;
else:
consonant += 1;
head=head.next;
print("Vowel: " , vowel , "");
print("Consonant: " , consonant , "");
if __name__ == '__main__':
head = Node('u');
head.next = Node('h');
head.next.next = Node('d');
head.next.next.next = Node('a');
head.next.next.next.next = Node('n');
count(head);
|
C#
using System;
public class GFG
{
class Node {
public char data;
public Node next;
public Node(char key)
{
data = key;
next = null;
}
};
static bool isVowel(char x)
{
return (x == 'a' || x == 'e'
|| x == 'i' || x == 'o'
|| x == 'u');
}
static void count(Node head)
{
int vowel = 0;
int consonant = 0;
for (Node itr = head; itr != null; itr = itr.next) {
if (isVowel(itr.data)) {
vowel++;
}
else {
consonant++;
}
}
Console.Write("Vowel: " + vowel +"\n");
Console.Write("Consonant: " + consonant +"\n");
}
public static void Main(String[] args)
{
Node head = new Node('u');
head.next = new Node('h');
head.next.next = new Node('d');
head.next.next.next = new Node('a');
head.next.next.next.next = new Node('n');
count(head);
}
}
|
Javascript
<script>
class Node
{
constructor(key)
{
this.data = key;
this.next = null;
}
};
function isVowel(x)
{
return (x == 'a' || x == 'e'
|| x == 'i' || x == 'o'
|| x == 'u');
}
function count(head) {
let vowel = 0;
let consonant = 0;
for (let itr = head; itr != null; itr = itr.next) {
if (isVowel(itr.data)) {
vowel++;
}
else {
consonant++;
}
}
document.write("Vowel: " + vowel + "<br>");
document.write("Consonant: " + consonant + "<br>");
}
let head = new Node('u');
head.next = new Node('h');
head.next.next = new Node('d');
head.next.next.next = new Node('a');
head.next.next.next.next = new Node('n');
count(head);
</script>
|
Output
Vowel: 2
Consonant: 3
Time Complexity: O(N)
Auxiliary Space: O(1)
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