# 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 ` `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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