Given a doubly-linked list containing N nodes and given a number K. The task is to find the sum of all such nodes which are divisible by K.
Examples:
Input: List = 15 <=> 16 <=> 10 <=> 9 <=> 6 <=> 7 <=> 17
K = 3
Output: Sum = 30
Input: List = 5 <=> 3 <=> 6 <=> 8 <=> 4 <=> 1 <=> 2 <=> 9
K = 2
Output: Sum = 20
Approach: 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 add that node value to otherwise continue this process while the end of the list is not reached.
Algorithm:
- To represent a doubly linked list, create a DLL class.
- Create the static method push() to add a new node to the list’s top.
- Using the provided data, create a new node.
- Set the new node’s next pointer to the list’s current head.
- Set the previous node’s reference to null.
- Set the previous pointer of the current head, if it is not null, to the new node.
- Position the head such that it faces the new node.
- Give the head back.
- Create a static function called sumOfNode() to add up all the doubly linked list nodes that are divided by K.
- Set a variable’s sum value to 0.
- Use a while loop to iterate through the list until the end is reached (node is null).
- If the current node’s data is divisible by K, include it in the sum.
- Go to the following node in the list.
- Return the total.
Below is the implementation of the above approach:
C++
// C++ implementation to add// all nodes value which is// divided by any given number K#include <bits/stdc++.h>using namespace std;// Node of the doubly linked liststruct 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 = (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 of 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 sum all the nodes// from the doubly linked// list that are divided by Kint sumOfNode(Node** head_ref, int K){ Node* ptr = *head_ref; Node* next; // variable sum=0 for add nodes value int sum = 0; // 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) sum += ptr->data; ptr = next; } // return sum of nodes which is divided by K return sum;}// Driver program to test aboveint 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 sum = sumOfNode(&head, 3); cout << "Sum = " << sum;} |
Java
// Java implementation to add// all nodes value which is// divided by any given number K// Node of the doubly linked listclass Node { int data; Node next, prev; Node(int d) { data = d; next = null; prev = null; }}class DLL { // function to insert a node at the beginning // of the Doubly Linked List static Node push(Node head, int data) { // allocate node Node newNode = new Node(data); newNode.next = head; // since we are adding at the beginning, // prev is always NULL newNode.prev = null; // change prev of head node to new node if (head != null) head.prev = newNode; // move the head to point to the new node head = newNode; return head; } // function to sum all the nodes // from the doubly linked // list that are divided by K static int sumOfNode(Node node, int K) { // variable sum=0 for add nodes value int sum = 0; // traverse list till last node while (node != null) { // check is node value divided by K // if true then add in sum if (node.data % K == 0) sum += node.data; node = node.next; } // return sum of nodes which is divided by K return sum; } // Driver program 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 sum = sumOfNode(head, 3); System.out.println("Sum = " + sum); }}// This code is contributed by Vivekkumar Singh |
Python3
# Python3 implementation to add# all nodes value which is# divided by any given number 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 adding at the beginning, # prev is always None new_node.prev = None # link the old list of 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 sum all the nodes# from the doubly linked# list that are divided by Kdef sumOfNode(head_ref, K): ptr = head_ref next = None # variable sum=0 for add nodes value sum = 0 # traverse list till last node while (ptr != None) : next = ptr.next # check is node value divided by K # if true then add in sum if (ptr.data % K == 0): sum += ptr.data ptr = next # return sum of nodes which is divided by K return sum# 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) sum = sumOfNode(head, 3) print("Sum =", sum)# This code is contributed by Arnab Kundu |
C#
// C# implementation to add // all nodes value which is // divided by any given number K using System;// Node of the doubly linked list public class Node { public int data; public Node next, prev; public Node(int d) { data = d; next = null; prev = null; } } class DLL { // function to insert a node at the beginning // of the Doubly Linked List static Node push(Node head, int data) { // allocate node Node newNode = new Node(data); newNode.next = head; // since we are adding at the beginning, // prev is always NULL newNode.prev = null; // change prev of head node to new node if (head != null) head.prev = newNode; // move the head to point to the new node head = newNode; return head; } // function to sum all the nodes // from the doubly linked // list that are divided by K static int sumOfNode(Node node, int K) { // variable sum=0 for add nodes value int sum = 0; // traverse list till last node while (node != null) { // check is node value divided by K // if true then add in sum if (node.data % K == 0) sum += node.data; node = node.next; } // return sum of nodes which is divided by K return sum; } // 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 sum = sumOfNode(head, 3); Console.WriteLine("Sum = " + sum); } } // This code is contributed by Arnab Kundu |
Javascript
<script>// JavaScript implementation to add// all nodes value which is// divided by any given number K// Node of the doubly linked listclass 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 , data) { // allocate node var newNode = new Node(data); newNode.next = head; // since we are adding at the beginning, // prev is always NULL newNode.prev = null; // change prev of head node to new node if (head != null) head.prev = newNode; // move the head to point to the new node head = newNode; return head; } // function to sum all the nodes // from the doubly linked // list that are divided by K function sumOfNode(node , K) { // variable sum=0 for add nodes value var sum = 0; // traverse list till last node while (node != null) { // check is node value divided by K // if true then add in sum if (node.data % K == 0) sum += node.data; node = node.next; } // return sum of nodes which is divided by K return sum; } // Driver program // 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 sum = sumOfNode(head, 3); document.write("Sum = " + sum);// This code contributed by umadevi9616</script> |
Output
Sum = 30
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Complexity: O(1) because using constant variables
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