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 List struct Node { int data; Node* next; }; // Function to insert a node at the // beginning of the singly Linked List void 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 number int 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 product 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) & 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 Code int 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 approach class GFG{ // Node of Linked List static class Node { int data; Node next; }; // Function to insert a node at the // beginning of the singly Linked List static 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 number static 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 product static 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 Code public 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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C#
// C# program for the above approach using System; class GFG{ // Node of Linked List class Node { public int data; public Node next; }; // Function to insert a node at the // beginning of the singly Linked List static 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 number static 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 product static 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 Code public 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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