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 nodestruct Node { int data; Node *prev, *next;}; // function to insert a node at the beginning// of the Doubly Linked Listvoid 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 Kint 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 Codeint 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 Listdef 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 Kdef 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 Codeif __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 nodeclass 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 nodevar 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 povar 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 listvar 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
Time Complexity: O(N), where N is the number of nodes.
Auxiliary Space: O(1)
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