# 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 = 9360

Input: 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` `} `

## 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 `

## 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 `

Output:

```Sum = 42
Product = 9360
```

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

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Improved By : 29AjayKumar, Rohit_ranjan