Remove all the Even Digit Sum Nodes from a Doubly Linked List

Given a Doubly linked list containing N nodes, the task is to remove all the nodes from the list which contains elements whose digit sum is even.

Examples:

Input: DLL = 18 <=> 15 <=> 8 <=> 9 <=> 14
Output: 18 <=> 9 <=> 14
Explanation:
The linked list contains :
18 -> 1 + 8 = 9
15 -> 1 + 5 = 6
8 -> 8
9 -> 9
14 -> 1 + 4 = 5
Here, digit sum for nodes containing 15 and 8 are even.
Hence, these nodes have been deleted.

Input: DLL = 5 <=> 3 <=> 4 <=> 2 <=> 9
Output: 5 <=> 3 <=> 9
Explanation:
The linked list contains two digit sum values 4 and 2.
Hence, these nodes have been deleted.

Approach:
A simple approach is to traverse the nodes of the doubly linked list one by one and for each node first, find the digit sum for the value present in the node by iterating through each digit and then finally remove the nodes whose digit sum is even.



Below is the implementation of the above approach:

C++

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// C++ implementation to remove all
// the Even Digit Sum Nodes from a
// doubly linked list
  
#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 the node
    Node* new_node
        = (Node*)malloc(sizeof(struct Node));
  
    // Insert 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 the 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 digit sum
// for a number
int digitSum(int num)
{
    int sum = 0;
    while (num) {
        sum += (num % 10);
        num /= 10;
    }
  
    return sum;
}
  
// Function to delete a node
// in a Doubly Linked List.
// head_ref --> pointer to head node pointer.
// del --> pointer to node to be deleted
void deleteNode(Node** head_ref, Node* del)
{
    // Base case
    if (*head_ref == NULL || del == NULL)
        return;
  
    // If the node to be deleted is head node
    if (*head_ref == del)
        *head_ref = del->next;
  
    // Change next only if node to be
    // deleted is not the last node
    if (del->next != NULL)
        del->next->prev = del->prev;
  
    // Change prev only if node to be
    // deleted is not the first node
    if (del->prev != NULL)
        del->prev->next = del->next;
  
    // Finally, free the memory
    // occupied by del
    free(del);
  
    return;
}
  
// Function to to remove all
// the Even Digit Sum Nodes from a
// doubly linked list
void deleteEvenDigitSumNodes(Node** head_ref)
{
    Node* ptr = *head_ref;
    Node* next;
  
    // Iterating through the linked list
    while (ptr != NULL) {
        next = ptr->next;
  
        // If node's data's digit sum
        // is even
        if (!(digitSum(ptr->data) & 1))
            deleteNode(head_ref, ptr);
  
        ptr = next;
    }
}
  
// Function to print nodes in a
// given doubly linked list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}
  
// Driver Code
int main()
{
  
    Node* head = NULL;
  
    // Create the doubly linked list
    // 18 <-> 15 <-> 8 <-> 9 <-> 14
    push(&head, 14);
    push(&head, 9);
    push(&head, 8);
    push(&head, 15);
    push(&head, 18);
  
    cout << "Original List: ";
    printList(head);
  
    deleteEvenDigitSumNodes(&head);
  
    cout << "\nModified List: ";
    printList(head);
}

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Output:

Original List: 18 15 8 9 14 
Modified List: 18 9 14

Time Complexity: O(N), where N is the total number of nodes.

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