Remove all the Even Digit Sum Nodes from a Circular Singly Linked List

Given a circular singly 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: CLL = 9 -> 11 -> 34 -> 6 -> 13 -> 21
Output: 9 -> 34 -> 21
Explanation:
The circular singly linked list contains :
9 -> 9
11 -> 1 + 1 = 2
34 -> 3 + 4 = 7
6 -> 6
13 -> 1 + 3 = 4
21 -> 2 + 1 = 3
Here, digit sum for nodes containing 11, 6 and 13 are even.
Hence, these nodes have been deleted.



Input: 5 -> 11 -> 16 -> 21 -> 17 -> 10
Output: 5 -> 16 -> 21 -> 10
Explanation:
The linked list contains two digit sum values 11 and 17.
Hence, these nodes have been deleted.

Approach: The idea is to traverse the nodes of the circular singly linked list one by one and for each node, find the digit sum for the value present in the node by iterating through each digit. If the digit sum is even, then remove the nodes. Else, continue.

Below is the implementation of the above approach:

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// C++ program to remove all
// the Even Digit Sum Nodes from a
// circular singly linked list
  
#include <bits/stdc++.h>
using namespace std;
  
// Structure for a node
struct Node {
    int data;
    struct Node* next;
};
  
// Function to insert a node at the beginning
// of a Circular linked list
void push(struct Node** head_ref, int data)
{
    // Create a new node and make head as next
    // of it.
    struct Node* ptr1
        = (struct Node*)malloc(
            sizeof(struct Node));
  
    struct Node* temp = *head_ref;
    ptr1->data = data;
    ptr1->next = *head_ref;
  
    // If linked list is not NULL then
    // set the next of last node
    if (*head_ref != NULL) {
  
        // Find the node before head
        // and update next of it.
        while (temp->next != *head_ref)
            temp = temp->next;
  
        temp->next = ptr1;
    }
    else
  
        // Point for the first node
        ptr1->next = ptr1;
  
    *head_ref = ptr1;
}
  
// Function to delete the node from a 
// Circular Linked list
void deleteNode(Node* head_ref, Node* del)
{
    struct Node* temp = head_ref;
  
    // If node to be deleted is head node
    if (head_ref == del)
        head_ref = del->next;
  
    // Traverse list till not found
    // delete node
    while (temp->next != del) {
        temp = temp->next;
    }
  
    // Copy the address of the node
    temp->next = del->next;
  
    // Finally, free the memory
    // occupied by del
    free(del);
  
    return;
}
  
// 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 all the Even Digit Sum Nodes
// from the singly circular linked list
void deleteEvenDigitSumNodes(Node* head)
{
    struct Node* ptr = head;
  
    struct Node* next;
  
    // Traverse the list till the end
    do {
  
        // If the node's data is Fibonacci,
        // delete node 'ptr'
        if (!(digitSum(ptr->data) & 1))
            deleteNode(head, ptr);
  
        // Point to the next node
        next = ptr->next;
        ptr = next;
  
    } while (ptr != head);
}
  
// Function to print nodes in a
// given Circular linked list
void printList(struct Node* head)
{
    struct Node* temp = head;
    if (head != NULL) {
        do {
            printf("%d ", temp->data);
            temp = temp->next;
        } while (temp != head);
    }
}
  
// Driver code
int main()
{
    // Initialize lists as empty
    struct Node* head = NULL;
  
    // Created linked list will be
    // 9->11->34->6->13->21
    push(&head, 21);
    push(&head, 13);
    push(&head, 6);
    push(&head, 34);
    push(&head, 11);
    push(&head, 9);
  
    deleteEvenDigitSumNodes(head);
  
    printList(head);
  
    return 0;
}

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Output:

9 34 21

Time Complexity: O(KN), where N is the size of the linked list and K is the number of digits in the maximum number of the linked list.

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