Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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

  • Difficulty Level : Easy
  • Last Updated : 29 Oct, 2021

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

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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.
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 list
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 sum all the nodes
// from the doubly linked
// list that are divided by K
int sumOfNode(Node** head_ref, int K)
{
    Node* ptr = *head_ref;
    Node* next;
    // variable sum=0 for add nodes value
    int sum = 0;
    // traves 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 above
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 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 list
class 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 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 adding 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 sum all the nodes
# from the doubly linked
# list that are divided by K
def 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 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)
   
    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 list
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 , 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

 

Time Complexity: O(N)
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :