Sum and Product of the nodes of a Singly Linked List which are divisible by K

Given a singly linked list. The task is to find the sum and product of all of the nodes of the given linked list which are divisible by a given number k.

Examples:

Input : List = 7->60->8->40->1
        k = 10 
Output : Product = 2400, Sum = 100
Product of nodes: 60 * 40 = 2400

Input : List = 15->7->3->9->11->5
        k = 5
Output : Product = 75, Sum = 20

Algorithm:



  1. Initialize a pointer ptr with the head of the linked list, a product variable with 1 and a sum variable with 0.
  2. Start traversing the linked list using a loop until all the nodes get traversed.
  3. For every node:
    • Multiply the value of the current node to the product if current node is divisible by k.
    • Add the value of the current node to the sum if current node is divisible by k.
  4. Increment the pointer to the next node of linked list i.e. ptr = ptr ->next.
  5. Repeat the above steps until end of linked list is reached.
  6. Finally, print the product and sum.

Below is the implementation of the above approach:

C++

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// C++ implementation to find the product
// and sum of nodes which are divisible by k
  
#include <iostream>
using namespace std;
  
// A Linked list node
struct Node {
    int data;
    struct Node* next;
};
  
// Function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = new Node;
  
    /* put in the data */
    new_node->data = new_data;
  
    /* link the old list to the new node */
    new_node->next = (*head_ref);
  
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
  
// Function to find the product and sum of
// nodes which are divisible by k
// of the given linked list
void productAndSum(struct Node* head, int k)
{
    // Pointer to traverse the list
    struct Node* ptr = head;
  
    int product = 1; // Variable to store product
    int sum = 0; // Variable to store sum
  
    // Traverse the list and
    // calculate the product
    // and sum
    while (ptr != NULL) {
        if (ptr->data % k == 0) {
            product *= ptr->data;
            sum += ptr->data;
        }
  
        ptr = ptr->next;
    }
  
    // Print the product and sum
    cout << "Product = " << product << endl;
    cout << "Sum = " << sum;
}
  
// Driver Code
int main()
{
    struct Node* head = NULL;
  
    // create linked list 70->6->8->4->10
    push(&head, 70);
    push(&head, 6);
    push(&head, 8);
    push(&head, 4);
    push(&head, 10);
  
    int k = 10;
  
    productAndSum(head, k);
  
    return 0;
}

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Java

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// Java implementation to find the product 
// and sum of nodes which are divisible by k 
class GFG 
{
      
// A Linked list node 
static class Node 
    int data; 
    Node next; 
}; 
  
// Function to insert a node at the 
// beginning of the linked list 
static Node push( Node head_ref, int new_data) 
    // allocate node /
    Node new_node = new Node(); 
  
    // put in the data /
    new_node.data = new_data; 
  
    // link the old list to the new node /
    new_node.next = (head_ref); 
  
    // move the head to point to the new node /
    (head_ref) = new_node; 
    return head_ref;
  
// Function to find the product and sum of 
// nodes which are divisible by k 
// of the given linked list 
static void productAndSum( Node head, int k) 
    // Pointer to traverse the list 
    Node ptr = head; 
  
    int product = 1; // Variable to store product 
    int sum = 0; // Variable to store sum 
  
    // Traverse the list and 
    // calculate the product 
    // and sum 
    while (ptr != null)
    
        if (ptr.data % k == 0)
        
            product *= ptr.data; 
            sum += ptr.data; 
        
  
        ptr = ptr.next; 
    
  
    // Print the product and sum 
    System.out.println( "Product = " + product ); 
    System.out.println( "Sum = " + sum); 
  
// Driver Code 
public static void main(String args[])
    Node head = null
  
    // create linked list 70.6.8.4.10 
    head = push(head, 70); 
    head = push(head, 6); 
    head = push(head, 8); 
    head = push(head, 4); 
    head = push(head, 10); 
  
    int k = 10
  
    productAndSum(head, k); 
  
}
}
  
// This code is contributed by Arnab Kundu

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C#

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// C# implementation to find the product 
// and sum of nodes which are divisible by k 
using System;
      
class GFG 
{
      
// A Linked list node 
public class Node 
    public int data; 
    public Node next; 
}; 
  
// Function to insert a node at the 
// beginning of the linked list 
static Node push( Node head_ref, int new_data) 
    // allocate node /
    Node new_node = new Node(); 
  
    // put in the data /
    new_node.data = new_data; 
  
    // link the old list to the new node /
    new_node.next = (head_ref); 
  
    // move the head to point to the new node /
    (head_ref) = new_node; 
    return head_ref;
  
// Function to find the product and sum of 
// nodes which are divisible by k 
// of the given linked list 
static void productAndSum( Node head, int k) 
    // Pointer to traverse the list 
    Node ptr = head; 
  
    int product = 1; // Variable to store product 
    int sum = 0; // Variable to store sum 
  
    // Traverse the list and 
    // calculate the product 
    // and sum 
    while (ptr != null)
    
        if (ptr.data % k == 0)
        
            product *= ptr.data; 
            sum += ptr.data; 
        
  
        ptr = ptr.next; 
    
  
    // Print the product and sum 
    Console.WriteLine( "Product = " + product ); 
    Console.WriteLine( "Sum = " + sum); 
  
// Driver Code 
public static void Main(String []args)
    Node head = null
  
    // create linked list 70.6.8.4.10 
    head = push(head, 70); 
    head = push(head, 6); 
    head = push(head, 8); 
    head = push(head, 4); 
    head = push(head, 10); 
  
    int k = 10; 
  
    productAndSum(head, k); 
  
}
}
  
/* This code contributed by PrinciRaj1992 */

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Output:

Product = 700
Sum = 80

Time Complexity: O(N), where N is the number of nodes in the linked list.



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