Product of all nodes in a doubly linked list divisible by a given number K

Given a doubly-linked list containing N nodes and given a number K. The task is to find the product of all such nodes which are divisible by K.

Examples:

Input : List = 15 <=> 16 <=> 10 <=> 9 <=> 6 <=> 7 <=> 17
        K = 3
Output : Product = 810

Input : List = 5 <=> 3 <=> 6 <=> 8 <=> 4 <=> 1 <=> 2 <=> 9
        K = 2
Output : Product = 384

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to traverse the doubly linked list and check the nodes one by one. If a node’s value is divisible by K then multiply that node’s value with the product so far and continue this process while the end of the list is not reached.

Below is the implementation of the above approach:

C++

// C++ program to find product of nodes in a 
// doubly linked list divisible by K 
  
#include <iostream> 
using namespace std; 
  
// Doubly linked list node 
struct Node { 
    int data; 
    Node *prev, *next; 
}; 
  
// function to insert a node at the beginning 
// of the Doubly Linked List 
void push(Node** head_ref, int new_data) 
{ 
    // allocate node 
    Node* new_node = (Node*)malloc(sizeof(struct Node)); 
  
    // put in the data 
    new_node->data = new_data; 
  
    // since we are adding at the beginning, 
    // prev is always NULL 
    new_node->prev = NULL; 
  
    // link the old list off the new node 
    new_node->next = (*head_ref); 
  
    // change prev of head node to new node 
    if ((*head_ref) != NULL) 
        (*head_ref)->prev = new_node; 
  
    // move the head to point to the new node 
    (*head_ref) = new_node; 
} 
  
// Function to find the product of all the nodes from 
// the doubly linked list that is divisible by K 
int productOfNode(Node** head_ref, int K) 
{ 
    Node* ptr = *head_ref; 
    Node* next; 
  
    int product = 1; 
  
    // Travese list till last node 
    while (ptr != NULL) { 
        next = ptr->next; 
        // check is node value divided by K 
        // if true then add in sum 
        if (ptr->data % K == 0) 
            product *= ptr->data; 
        ptr = next; 
    } 
  
    // Return product of nodes which 
    // are divisible by K 
    return product; 
} 
  
// Driver Code 
int main() 
{ 
    // start with the empty list 
    Node* head = NULL; 
  
    // create the doubly linked list 
    // 15 16 10 9 6 7 17 
    push(&head, 17); 
    push(&head, 7); 
    push(&head, 6); 
    push(&head, 9); 
    push(&head, 10); 
    push(&head, 16); 
    push(&head, 15); 
  
    int K = 3; 
  
    int prod = productOfNode(&head, K); 
  
    cout << "Product = " << prod; 
  
    return 0; 
} 

Java

// Java program to find product of nodes in a  
// doubly linked list divisible by K  
class GFG 
{ 
  
// Doubly linked list node  
static class Node 
{  
    int data;  
    Node prev, next;  
};  
  
// function to insert a node at the beginning  
// of the Doubly Linked List  
static Node push(Node head_ref, int new_data)  
{  
    // allocate node  
    Node new_node = new Node();  
  
    // put in the data  
    new_node.data = new_data;  
  
    // since we are adding at the beginning,  
    // prev is always null  
    new_node.prev = null;  
  
    // link the old list off the new node  
    new_node.next = (head_ref);  
  
    // change prev of head node to new node  
    if ((head_ref) != null)  
        (head_ref).prev = new_node;  
  
    // move the head to point to the new node  
    (head_ref) = new_node;  
    return head_ref; 
}  
  
// Function to find product of all the nodes from  
// the doubly linked list that are divisible by K  
static int productOfNode(Node head_ref, int K)  
{  
    Node ptr = head_ref;  
    Node next;  
  
    int product = 1;  
  
    // Travese list till last node  
    while (ptr != null) 
    {  
        next = ptr.next;  
          
        // check is node value divided by K  
        // if true then add in sum  
        if (ptr.data % K == 0)  
            product *= ptr.data;  
        ptr = next;  
    }  
  
    // Return product of nodes which  
    // are divisible by K  
    return product;  
}  
  
// Driver Code  
public static void main(String args[]) 
{  
    // start with the empty list  
    Node head = null;  
  
    // create the doubly linked list  
    // 15 16 10 9 6 7 17  
    head = push(head, 17);  
    head = push(head, 7);  
    head = push(head, 6);  
    head = push(head, 9);  
    head = push(head, 10);  
    head = push(head, 16);  
    head = push(head, 15);  
  
    int K = 3;  
  
    int prod = productOfNode(head, K);  
  
    System.out.println( "Product = " + prod);  
} 
}  
  
// This code is contributed by Arnab Kundu 

Python3

# Python3 program to find product of nodes in a 
# doubly linked list divisible by K 
  
# Node of the doubly linked list  
class Node:  
      
    def __init__(self, data):  
        self.data = data  
        self.prev = None
        self.next = None
  
# function to insert a node at the beginning 
# of the Doubly Linked List 
def push(head_ref, new_data): 
  
    # allocate node 
    new_node = Node(0) 
  
    # put in the data 
    new_node.data = new_data 
  
    # since we are multiplying at the beginning, 
    # prev is always None 
    new_node.prev = None
  
    # link the old list off the new node 
    new_node.next = (head_ref) 
  
    # change prev of head node to new node 
    if ((head_ref) != None): 
        (head_ref).prev = new_node 
  
    # move the head to point to the new node 
    (head_ref) = new_node 
    return head_ref 
  
# function to product all the nodes 
# from the doubly linked 
# list that are divided by K 
def productOfNode(head_ref, K): 
  
    ptr = head_ref 
    next = None
      
    # variable product=1  
    product = 1
      
    # traves list till last node 
    while (ptr != None) : 
        next = ptr.next
          
        # check is node value divided by K 
        # if true then multiply in product 
        if (ptr.data % K == 0): 
            product *= ptr.data 
        ptr = next
      
    # return product of nodes which is divided by K 
    return product 
  
# Driver Code 
if __name__ == "__main__":  
  
    # start with the empty list 
    head = None
  
    # create the doubly linked list 
    # 15 <. 16 <. 10 <. 9 <. 6 <. 7 <. 17 
    head = push(head, 17) 
    head = push(head, 7) 
    head = push(head, 6) 
    head = push(head, 9) 
    head = push(head, 10) 
    head = push(head, 16) 
    head = push(head, 15) 
      
    K = 3
      
    product = productOfNode(head, K) 
    print("product =", product) 
  
# This code is contributed by Arnab Kundu 

C#

// C# program to find product of nodes in a  
// doubly linked list divisible by K  
using System; 
  
class GFG 
{ 
  
// Doubly linked list node  
public class Node 
{  
    public int data;  
    public Node prev, next;  
};  
  
// function to insert a node at the beginning  
// of the Doubly Linked List  
static Node push(Node head_ref, int new_data)  
{  
    // allocate node  
    Node new_node = new Node();  
  
    // put in the data  
    new_node.data = new_data;  
  
    // since we are adding at the beginning,  
    // prev is always null  
    new_node.prev = null;  
  
    // link the old list off the new node  
    new_node.next = (head_ref);  
  
    // change prev of head node to new node  
    if ((head_ref) != null)  
        (head_ref).prev = new_node;  
  
    // move the head to point to the new node  
    (head_ref) = new_node;  
    return head_ref; 
}  
  
// Function to find product of all the nodes from  
// the doubly linked list that are divisible by K  
static int productOfNode(Node head_ref, int K)  
{  
    Node ptr = head_ref;  
    Node next;  
  
    int product = 1;  
  
    // Travese list till last node  
    while (ptr != null) 
    {  
        next = ptr.next;  
          
        // check is node value divided by K  
        // if true then add in sum  
        if (ptr.data % K == 0)  
            product *= ptr.data;  
        ptr = next;  
    }  
  
    // Return product of nodes which  
    // are divisible by K  
    return product;  
}  
  
// Driver Code  
public static void Main(String []args) 
{  
    // start with the empty list  
    Node head = null;  
  
    // create the doubly linked list  
    // 15 16 10 9 6 7 17  
    head = push(head, 17);  
    head = push(head, 7);  
    head = push(head, 6);  
    head = push(head, 9);  
    head = push(head, 10);  
    head = push(head, 16);  
    head = push(head, 15);  
  
    int K = 3;  
  
    int prod = productOfNode(head, K);  
  
    Console.WriteLine( "Product = " + prod);  
} 
} 
  
// This code contributed by Rajput-Ji 

Output:

Product = 810

Time Complexity: O(N), where N is the number of nodes.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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