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: 3Input: 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++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// A linked list nodestruct Node { char data; struct Node* next; Node(char key) { data = key; next = NULL; }};// Utility function to check if a character// is a vowel or notbool isVowel(char x){ return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u');}// Function to count the number// of vowels and consonantsvoid 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;}// Driver Codeint 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
// Java program for the above approachimport java.util.*;class GFG{ // A linked list node static class Node { char data; Node next; Node(char key) { data = key; next = null; } }; // Utility function to check if a character // is a vowel or not static boolean isVowel(char x) { return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u'); } // Function to count the number // of vowels and consonants 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"); } // Driver Code 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); }}// This code is contributed by Rajput-Ji |
Python3
# Python program for the above approach# A linked list Nodeclass Node: def __init__(self, data): self.data = data; self.next = None;# Utility function to check if a character# is a vowel or notdef isVowel(x): return (x == 'a' or x == 'e' or x == 'i' or x == 'o' or x == 'u');# Function to count the number# of vowels and consonantsdef 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 , "");# Driver Codeif __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);# This code is contributed by 29AjayKumar |
C#
// C# program for the above approachusing System;public class GFG{ // A linked list node class Node { public char data; public Node next; public Node(char key) { data = key; next = null; } }; // Utility function to check if a character // is a vowel or not static bool isVowel(char x) { return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u'); } // Function to count the number // of vowels and consonants 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"); } // Driver Code 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); }}// This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript code for the above approach // A linked list node class Node { constructor(key) { this.data = key; this.next = null; } }; // Utility function to check if a character // is a vowel or not function isVowel(x) { return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u'); } // Function to count the number // of vowels and consonants 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>"); } // Driver Code 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); // This code is contributed by Potta Lokesh </script> |
Output
Vowel: 2 Consonant: 3
Time Complexity: O(N)
Auxiliary Space: O(1)
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