Count of Prime Nodes of a Singly Linked List
Last Updated :
07 Dec, 2023
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 list
struct Node {
int data;
Node* next;
};
// Function to insert a node at the beginning
// of the singly Linked List
void 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 prime
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
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
int 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 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 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 node
class 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 Code
if __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# 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
<script>
// Javascript implementation to find count
// of prime numbers in the singly linked list
// Node of the singly linked list
class Node
{
constructor(val)
{
this.data = val;
this.next = null;
}
}
// Function to insert a node at the beginning
// of the singly Linked List
function 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 prime
function 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 list
function 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 list
var 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
document.write("Count of prime nodes = " +
countPrime(head));
// This code is contributed by gauravrajput1
</script>
OutputCount 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 list
struct Node {
int data;
Node* next;
};
// Function to insert a node at the beginning
// of the singly linked list
void 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 prime
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 count prime nodes in a linked list
int countPrimeRecursive(Node* head) {
if (head == NULL)
return 0;
int count = countPrimeRecursive(head->next);
if (isPrime(head->data))
count++;
return count;
}
// Driver program
int 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;
}
// Java program of the above approach
class Node {
int data;
Node next;
Node(int data)
{
this.data = data;
this.next = null;
}
}
public class GFG {
// Function to insert a node at the beginning
// of the singly linked list
static Node push(Node head, int newData)
{
Node newNode = new Node(newData);
newNode.next = head;
return newNode;
}
// 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 count prime nodes in a linked list
static int countPrimeRecursive(Node head)
{
if (head == null)
return 0;
int count = countPrimeRecursive(head.next);
if (isPrime(head.data))
count++;
return count;
}
// Driver program
public static void main(String[] args)
{
// Start with an 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 the required answer
System.out.println("Count of prime nodes = "
+ countPrimeRecursive(head));
}
}
// This code is contributed by Susobhan Akhuli
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Function to insert a node at the beginning
# of the singly linked list
def push(head, new_data):
new_node = Node(new_data)
new_node.next = head
return new_node
# Function to check if a number is prime
def is_prime(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
# Function to count prime nodes in a linked list
def count_prime_recursive(head):
if head is None:
return 0
count = count_prime_recursive(head.next)
if is_prime(head.data):
count += 1
return count
# Driver program
if __name__ == "__main__":
# Start with an empty list
head = None
# 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 the required answer
print("Count of prime nodes =", count_prime_recursive(head))
using System;
// Node of the singly linked list
class Node
{
public int data;
public Node next;
public Node(int data)
{
this.data = data;
this.next = null;
}
}
public class GFG
{
// Function to insert a node at the beginning of the singly linked list
static Node Push(Node head, int newData)
{
Node newNode = new Node(newData);
newNode.next = head;
return newNode;
}
// 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 the loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i += 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to count prime nodes in a linked list
static int CountPrimeRecursive(Node head)
{
if (head == null)
return 0;
int count = CountPrimeRecursive(head.next);
if (IsPrime(head.data))
count++;
return count;
}
// Driver program
public static void Main(string[] args)
{
// Start with an 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 count prime nodes and print the result
Console.WriteLine("Count of prime nodes = " + CountPrimeRecursive(head));
}
}
// Node of the singly linked list
class Node {
constructor(data) {
this.data = data;
this.next = null;
}
}
// Function to insert a node at the beginning
// of the singly linked list
function push(head, new_data) {
let new_node = new Node(new_data);
new_node.next = head;
head = new_node;
return head;
}
// Function to check if a number is prime
function 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 (let 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 list
function countPrimeRecursive(head) {
if (head === null)
return 0;
let count = countPrimeRecursive(head.next);
if (isPrime(head.data))
count++;
return count;
}
// Driver program
let 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 the required answer
console.log("Count of prime nodes =", countPrimeRecursive(head));
OutputCount 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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