Given a singly linked list containing N nodes, the task is to find the total count of prime numbers.
Examples:
Input: List = 15 -> 5 -> 6 -> 10 -> 17 Output: 2 5 and 17 are the prime nodes Input: List = 29 -> 3 -> 4 -> 2 -> 9 Output: 3 2, 3 and 29 are the prime nodes
Approach: The idea is to traverse the linked list to the end and check if the current node is prime or not. If YES, increment the count by 1 and keep doing the same until all the nodes get traversed.
Below is the implementation of above approach:
C++
// C++ implementation to find count of prime numbers// in the singly linked list#include <bits/stdc++.h>using namespace std;// Node of the singly linked liststruct Node { int data; Node* next;};// Function to insert a node at the beginning// of the singly Linked Listvoid push(Node** head_ref, int new_data){ Node* new_node = new Node; new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node;}// Function to check if a number is primebool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to find count of prime// nodes in a linked listint countPrime(Node** head_ref){ int count = 0; Node* ptr = *head_ref; while (ptr != NULL) { // If current node is prime if (isPrime(ptr->data)) { // Update count count++; } ptr = ptr->next; } return count;}// Driver programint main(){ // start with the empty list Node* head = NULL; // create the linked list // 15 -> 5 -> 6 -> 10 -> 17 push(&head, 17); push(&head, 10); push(&head, 6); push(&head, 5); push(&head, 15); // Function call to print require answer cout << "Count of prime nodes = " << countPrime(&head); return 0;} |
Java
// Java implementation to find count of prime numbers // in the singly linked list class solution{// Node of the singly linked list static class Node { int data; Node next; }// Function to insert a node at the beginning // of the singly Linked List static Node push(Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = ( head_ref); ( head_ref) = new_node; return head_ref;} // Function to check if a number is prime static boolean isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // Function to find count of prime // nodes in a linked list static int countPrime(Node head_ref) { int count = 0; Node ptr = head_ref; while (ptr != null) { // If current node is prime if (isPrime(ptr.data)) { // Update count count++; } ptr = ptr.next; } return count; } // Driver program public static void main(String args[]){ // start with the empty list Node head = null; // create the linked list // 15 . 5 . 6 . 10 . 17 head=push(head, 17); head=push(head, 10); head=push(head, 6); head=push(head, 5); head=push(head, 15); // Function call to print require answer System.out.print( "Count of prime nodes = "+ countPrime(head)); } }// This code is contributed by Arnab Kundu |
Python3
# Python3 implementation to find count of # prime numbers in the singly linked list# Function to check if a number is prime def isPrime(n): # Corner cases if n <= 1: return False if n <= 3: return True # This is checked so that we can skip # middle five numbers in below loop if n % 2 == 0 or n % 3 == 0: return False i = 5 while i * i <= n: if n % i == 0 or n % (i + 2) == 0: return False i += 6 return True# Link list nodeclass Node: def __init__(self, data, next): self.data = data self.next = next class LinkedList: def __init__(self): self.head = None # Push a new node on the front of the list. def push(self, new_data): new_node = Node(new_data, self.head) self.head = new_node # Function to find count of prime # nodes in a linked list def countPrime(self): count = 0 ptr = self.head while ptr != None: # If current node is prime if isPrime(ptr.data): # Update count count += 1 ptr = ptr.next return count # Driver Codeif __name__ == "__main__": # Start with the empty list linkedlist = LinkedList() # create the linked list # 15 -> 5 -> 6 -> 10 -> 17 linkedlist.push(17) linkedlist.push(10) linkedlist.push(6) linkedlist.push(5) linkedlist.push(15) # Function call to print require answer print("Count of prime nodes =", linkedlist.countPrime()) # This code is contributed by Rituraj Jain |
C#
// C# implementation to find count of prime numbers // in the singly linked list using System;class GFG{// Node of the singly linked list public class Node { public int data; public Node next; }// Function to insert a node at the beginning // of the singly Linked List static Node push(Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = ( head_ref); ( head_ref) = new_node; return head_ref;} // Function to check if a number is prime static bool isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // Function to find count of prime // nodes in a linked list static int countPrime(Node head_ref) { int count = 0; Node ptr = head_ref; while (ptr != null) { // If current node is prime if (isPrime(ptr.data)) { // Update count count++; } ptr = ptr.next; } return count; } // Driver code public static void Main(String []args){ // start with the empty list Node head = null; // create the linked list // 15 . 5 . 6 . 10 . 17 head=push(head, 17); head=push(head, 10); head=push(head, 6); head=push(head, 5); head=push(head, 15); // Function call to print require answer Console.Write( "Count of prime nodes = "+ countPrime(head)); } }// This code has been contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation to find count // of prime numbers in the singly linked list // Node of the singly linked listclass Node { constructor(val) { this.data = val; this.next = null; }}// Function to insert a node at the beginning// of the singly Linked Listfunction push(head_ref, new_data) { var new_node = new Node(); new_node.data = new_data; new_node.next = (head_ref); (head_ref) = new_node; return head_ref;}// Function to check if a number is primefunction isPrime(n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for(i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to find count of prime// nodes in a linked listfunction countPrime(head_ref){ var count = 0; var ptr = head_ref; while (ptr != null) { // If current node is prime if (isPrime(ptr.data)) { // Update count count++; } ptr = ptr.next; } return count;}// Driver code// Start with the empty listvar head = null;// Create the linked list// 15 . 5 . 6 . 10 . 17head = push(head, 17);head = push(head, 10);head = push(head, 6);head = push(head, 5);head = push(head, 15);// Function call to print require answerdocument.write("Count of prime nodes = " + countPrime(head));// This code is contributed by gauravrajput1</script> |
Output
Count of prime nodes = 2
Complexity Analysis:
- Time Complexity: O(N*sqrt(P)), where N is length of the LinkedList and P is the maximum element in the List
- Auxiliary Space: O(1)
Recursive Approach:
The base case of the recursion is when the head node is NULL, in which case the function returns 0. Otherwise, the function first calls itself recursively for the next node in the linked list, and obtains the count of prime nodes in the remaining linked list. If the data of the current node is prime, it adds 1 to the count and returns it, otherwise it simply returns the count obtained from the recursive call. The final result returned by the function is the count of prime nodes in the entire linked list.
- If the head node is NULL, return 0.
- Recursively call the function for the rest of the linked list by passing the next node.
- If the data of the current node is prime, add 1 to the count and return it.
- Otherwise, return the count obtained from the recursive call.
Below is the implementation of the above approach:
C++
#include <iostream>using namespace std;// Node of the singly linked liststruct Node { int data; Node* next;};// Function to insert a node at the beginning// of the singly linked listvoid push(Node** head_ref, int new_data) { Node* new_node = new Node; new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node;}// Function to check if a number is primebool isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Function to count prime nodes in a linked listint countPrimeRecursive(Node* head) { if (head == NULL) return 0; int count = countPrimeRecursive(head->next); if (isPrime(head->data)) count++; return count;}// Driver programint main() { // start with the empty list Node* head = NULL; // create the linked list // 15 -> 5 -> 6 -> 10 -> 17 push(&head, 17); push(&head, 10); push(&head, 6); push(&head, 5); push(&head, 15); // Function call to print required answer cout << "Count of prime nodes = " << countPrimeRecursive(head); return 0;} |
Output
Count of prime nodes = 2
Time Complexity: O(N), where N is the number of nodes in the linked list.
Space Complexity: O(N), where N is the number of nodes in the linked list. This is because we create a recursive call stack for each node.
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