Sum and Product of all even digit sum Nodes of a Singly Linked List
Given a singly linked list containing N nodes, the task is to find the sum and product of all the nodes from the list whose data value has an even digit sum.
Examples:
Input: 15 -> 16 -> 8 -> 6 -> 13
Output:
Sum = 42
Product = 9360
Explanation:
The sum of all digit of number in linked list are:
15 = 1 + 5 = 6
16 = 1 + 6 = 7
8 = 8
6 = 6
13 = 1 + 3 = 4
The list contains 4 Even Digit Sum data values 15, 8, 6 and 13.
Sum = 15 + 8 + 6 + 13 = 42
Product = 15 * 8 * 6 * 13 = 9360Input: 5 -> 3 -> 4 -> 2 -> 9
Output:
Sum = 6
Product = 8
Explanation:
The list contains 2 Even Digit Sum data values 4 and 2.
Sum = 4 + 2 = 6
Product = 4 * 2 = 8
Approach: The idea is to traverse the given linked list and check whether the sum of all the digits of current node value is even or not. If yes include the current node value to the resultant sum and the resultant product Else check for the next node value.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Node of Linked Liststruct Node { int data; Node* next;};// Function to insert a node at the// beginning of the singly Linked Listvoid push(Node** head_ref, int new_data){ // Allocate new node Node* new_node = (Node*)malloc( sizeof(struct Node)); // Insert the data new_node->data = new_data; // Link old list to the new node new_node->next = (*head_ref); // Move head to point the new node (*head_ref) = new_node;}// Function to find the digit sum// for a numberint digitSum(int num){ int sum = 0; while (num) { sum += (num % 10); num /= 10; } // Return the sum return sum;}// Function to find the required// sum and productvoid sumAndProduct(Node* head_ref){ // Initialise the sum and product // to 0 and 1 respectively int prod = 1; int sum = 0; Node* ptr = head_ref; // Traverse the given linked list while (ptr != NULL) { // If current node has even // digit sum then include it in // resultant sum and product if (!(digitSum(ptr->data) & 1)) { // Find the sum and the product prod *= ptr->data; sum += ptr->data; } ptr = ptr->next; } // Print the final Sum and Product cout << "Sum = " << sum << endl; cout << "Product = " << prod;}// Driver Codeint main(){ // Head of the linked list Node* head = NULL; // Create the linked list // 15 -> 16 -> 8 -> 6 -> 13 push(&head, 13); push(&head, 6); push(&head, 8); push(&head, 16); push(&head, 15); // Function call sumAndProduct(head); return 0;} |
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Java
// Java program for the above approachclass GFG{// Node of Linked Liststatic class Node { int data; Node next;};// Function to insert a node at the// beginning of the singly Linked Liststatic Node push(Node head_ref, int new_data){ // Allocate new node Node new_node = new Node(); // Insert the data new_node.data = new_data; // Link old list to the new node new_node.next = head_ref; // Move head to point the new node head_ref = new_node; return head_ref;}// Function to find the digit sum// for a numberstatic int digitSum(int num){ int sum = 0; while (num > 0) { sum += (num % 10); num /= 10; } // Return the sum return sum;}// Function to find the required// sum and productstatic void sumAndProduct(Node head_ref){ // Initialise the sum and product // to 0 and 1 respectively int prod = 1; int sum = 0; Node ptr = head_ref; // Traverse the given linked list while (ptr != null) { // If current node has even // digit sum then include it in // resultant sum and product if ((digitSum(ptr.data) %2 != 1)) { // Find the sum and the product prod *= ptr.data; sum += ptr.data; } ptr = ptr.next; } // Print the final Sum and Product System.out.print("Sum = " + sum +"\n"); System.out.print("Product = " + prod);}// Driver Codepublic static void main(String[] args){ // Head of the linked list Node head = null; // Create the linked list // 15.16.8.6.13 head = push(head, 13); head = push(head, 6); head = push(head, 8); head = push(head, 16); head = push(head, 15); // Function call sumAndProduct(head);}}// This code is contributed by 29AjayKumar |
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Python3
# Python3 program for the above approach# Node of Linked Listclass Node: def __init__(self, x): self.data = x self.next = None# Function to insert a node at the# beginning of the singly Linked Listdef push(head_ref, new_data): # Insert the data new_node = Node(new_data) # Link old list to the new node new_node.next = head_ref # Move head to pothe new node head_ref = new_node return head_ref# Function to find the digit sum# for a numberdef digitSum(num): sum = 0 while (num): sum += (num % 10) num //= 10 # Return the sum return sum# Function to find the required# sum and productdef sumAndProduct(head_ref): # Initialise the sum and product # to 0 and 1 respectively prod = 1 sum = 0 ptr = head_ref # Traverse the given linked list while (ptr != None): # If current node has even # digit sum then include it in # resultant sum and product if (not (digitSum(ptr.data) & 1)): # Find the sum and the product prod *= ptr.data sum += ptr.data ptr = ptr.next # Print the final Sum and Product print("Sum =", sum) print("Product =", prod)# Driver Codeif __name__ == '__main__': # Head of the linked list head = None # Create the linked list # 15 . 16 . 8 . 6 . 13 head = push(head, 13) head = push(head, 6) head = push(head, 8) head = push(head, 16) head = push(head, 15) # Function call sumAndProduct(head) # This code is contributed by mohit kumar 29 |
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C#
// C# program for the above approachusing System;class GFG{// Node of Linked Listclass Node{ public int data; public Node next;};// Function to insert a node at the// beginning of the singly Linked Liststatic Node push(Node head_ref, int new_data){ // Allocate new node Node new_node = new Node(); // Insert the data new_node.data = new_data; // Link old list to the new node new_node.next = head_ref; // Move head to point the new node head_ref = new_node; return head_ref;}// Function to find the digit sum// for a numberstatic int digitSum(int num){ int sum = 0; while (num > 0) { sum += (num % 10); num /= 10; } // Return the sum return sum;}// Function to find the required// sum and productstatic void sumAndProduct(Node head_ref){ // Initialise the sum and product // to 0 and 1 respectively int prod = 1; int sum = 0; Node ptr = head_ref; // Traverse the given linked list while (ptr != null) { // If current node has even // digit sum then include it in // resultant sum and product if ((digitSum(ptr.data) % 2 != 1)) { // Find the sum and the product prod *= ptr.data; sum += ptr.data; } ptr = ptr.next; } // Print the readonly Sum and Product Console.Write("Sum = " + sum + "\n"); Console.Write("Product = " + prod);}// Driver Codepublic static void Main(String[] args){ // Head of the linked list Node head = null; // Create the linked list // 15.16.8.6.13 head = push(head, 13); head = push(head, 6); head = push(head, 8); head = push(head, 16); head = push(head, 15); // Function call sumAndProduct(head);}}// This code is contributed by Rohit_ranjan |
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Output:
Sum = 42 Product = 9360
Time Complexity: O(N), where N is the number of nodes in the linked list.
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