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

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 = 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;
}

// 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;

    // Traverse 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;

    // Traverse 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

    # traverse 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;

    // Traverse 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

Javascript

<script>
// javascript program to find product of nodes in a
// doubly linked list divisible by K
// Doubly linked list node
class Node {
    constructor(val) {
        this.data = val;
        this.prev = null;
        this.next = null;
    }
}

    // function to insert a node at the beginning
    // of the Doubly Linked List
    function push(head_ref , new_data) {
        // allocate node
var 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
    function productOfNode(head_ref , K) {
var ptr = head_ref;
var next;

        var product = 1;

        // Traverse 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

        // start with the empty list
var 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);

        var K = 3;

        var prod = productOfNode(head, K);

        document.write("Product = " + prod);

// This code contributed by umadevi9616
</script>

Output

Product = 810

Complexity Analysis:

• Time Complexity: O(N), where N is the number of nodes.
• Auxiliary Space: O(1)