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 node struct Node { char data; struct Node* next; Node( char key) { data = key; next = NULL; } }; // Utility function to check if a character // is a vowel or not bool isVowel( char x) { return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u' ); } // Function to count the number // of vowels and consonants 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; } // Driver Code 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
// Java program for the above approach import 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 Node class Node: def __init__( self , data): self .data = data; self . next = None ; # Utility function to check if a character # is a vowel or not def 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 consonants 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 , ""); # Driver Code 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); # This code is contributed by 29AjayKumar |
C#
// C# program for the above approach using 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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