Sum 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 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 sum otherwise continue this process while the end of the list is not reached.

Below is the implementation of the above approach:

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// 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 begining,
    // 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;
}

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Output:

Sum = 30

Time Complexity: O(N)



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Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

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