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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Python3














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# Python3 implementation to find the product 


# and sum of nodes which are divisible by k 


import math 


  


# A Linked list node 


class Node:  


    def __init__(self, data):  


        self.data = data  


        self.next = None


  


# Function to insert a node at the 


# beginning of the linked list 


def push(head_ref, new_data): 


      


    # allocate node  


    new_node =Node(new_data) 


  


    # 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 


def productAndSum(head, k): 


      


    # Pointer to traverse the list 


    ptr = head 


  


    product = 1 # Variable to store product 


    add = 0 # Variable to store sum 


  


    # Traverse the list and 


    # calculate the product 


    # and sum 


    while (ptr != None): 


        if (ptr.data % k == 0): 


            product = product * ptr.data 


            add = add + ptr.data 


      


        ptr = ptr.next


      


    # Print the product and sum 


    print("Product =", product) 


    print("Sum =", add) 


  


# Driver Code 


if __name__=='__main__':  


  


    head = None


  


    # 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) 


  


    k = 10


  


    productAndSum(head, k) 


  


# This code is contributed by Srathore 



















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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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