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; } |
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 |
Python3
# Python3 program for the above approach # Node of Linked List class Node: def __init__( self , x): self .data = x self . next = None # Function to insert a node at the # beginning of the singly Linked List def 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 point the new node head_ref = new_node return head_ref # Function to find the digit sum # for a number def digitSum(num): sum = 0 while (num): sum + = (num % 10 ) num / / = 10 # Return the sum return sum # Function to find the required # sum and product def 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 Code if __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 |
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 |
Javascript
<script> // javascript program for the above approach // Node of Linked List class Node { constructor(val) { this .data = val; this .next = null ; } } // Function to insert a node at the // beginning of the singly Linked List function push(head_ref , new_data) { // Allocate new node var 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 function digitSum(num) { var sum = 0; while (num > 0) { sum += (num % 10); num = parseInt(num/10); } // Return the sum return sum; } // Function to find the required // sum and product function sumAndProduct(head_ref) { // Initialise the sum and product // to 0 and 1 respectively var prod = 1; var sum = 0; var 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 document.write( "Sum = " + sum + "<br/>" ); document.write( "Product = " + prod); } // Driver Code // Head of the linked list var 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 contributed by gauravrajput1 </script> |
Sum = 42 Product = 9360
Time Complexity: O(N), where N is the number of nodes in the linked list.
Auxiliary Space: O(1)
Recursive Approach:
- Create a recursive function that takes a pointer to the head of the linked list, the sum and the product as arguments.
- If the head pointer is NULL, return the sum and the product.
- Calculate the digit sum of the data of the current node.
- If the digit sum is even, update the sum and the product accordingly.
- Recursively call the function with the next node in the linked list, the updated sum, and the updated product.
- Return the final sum and product.
Below is the implementation of the above approach:
C++
#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 = 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; } // 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; } // Recursive function to find the sum and product // of nodes with even digit sum in a linked list void sumAndProductHelper(Node* ptr, int & prod, int & sum) { // Base case: if the current node is null, return if (ptr == NULL) { return ; } // Recursive call on the next node sumAndProductHelper(ptr->next, prod, sum); // If the current node has even digit sum, // update the product and sum if (!(digitSum(ptr->data) & 1)) { prod *= ptr->data; sum += ptr->data; } } // Wrapper function to call the recursive function void sumAndProduct(Node* head_ref) { // Initialise the sum and product to 0 and 1 respectively int prod = 1; int sum = 0; // Call the recursive function on the head node sumAndProductHelper(head_ref, prod, sum); // 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
// Node class for Linked List class Node { int data; Node next; Node( int data) { this .data = data; this .next = null ; } } public class Main { // Function to insert a node at the // beginning of the singly Linked List static Node push(Node head, int newData) { // Allocate new node Node newNode = new Node(newData); // Link old list to the new node newNode.next = head; // Move head to point to the new node return newNode; } // 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; } // Recursive function to find the sum and product // of nodes with even digit sum in a linked list static void sumAndProductHelper(Node ptr, int [] prodSum) { // Base case: if the current node is null, return if (ptr == null ) { return ; } // Recursive call on the next node sumAndProductHelper(ptr.next, prodSum); // If the current node has even digit sum, // update the product and sum if ((digitSum(ptr.data) & 1 ) == 0 ) { prodSum[ 0 ] *= ptr.data; prodSum[ 1 ] += ptr.data; } } // Wrapper function to call the recursive function static void sumAndProduct(Node head) { // Initialize the sum and product to 0 and 1 respectively int [] prodSum = { 1 , 0 }; // Call the recursive function on the head node sumAndProductHelper(head, prodSum); // Print the final Sum and Product System.out.println( "Sum = " + prodSum[ 1 ]); System.out.println( "Product = " + prodSum[ 0 ]); } // 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); } } |
Python
# Node of Linked List class Node: def __init__( self , data): self .data = data self . next = None # Function to insert a node at the # beginning of a singly Linked List def push(head_ref, new_data): # Create a new node new_node = Node(new_data) # Link old list to the new node new_node. next = head_ref[ 0 ] # Move head to point to the new node head_ref[ 0 ] = new_node # Function to find the digit sum for a number def digit_sum(num): _sum = 0 while num: _sum + = num % 10 num / / = 10 # Return the sum return _sum # Recursive function to find the sum and product # of nodes with even digit sum in a linked list def sum_and_product_helper(ptr, prod, _sum): # Base case: if the current node is None, return if ptr is None : return # Recursive call on the next node sum_and_product_helper(ptr. next , prod, _sum) # If the current node has an even digit sum, # update the product and sum if not digit_sum(ptr.data) & 1 : prod[ 0 ] * = ptr.data _sum[ 0 ] + = ptr.data # Wrapper function to call the recursive function def sum_and_product(head_ref): # Initialize the sum and product to 0 and 1, respectively prod = [ 1 ] _sum = [ 0 ] # Call the recursive function on the head node sum_and_product_helper(head_ref[ 0 ], prod, _sum) # Print the final Sum and Product print ( "Sum =" , _sum[ 0 ]) print ( "Product =" , prod[ 0 ]) # Driver Code if __name__ = = "__main__" : # Head of the linked list head = [ None ] # 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 sum_and_product(head) |
C#
using System; // Node of Linked List public class Node { public int data; public Node next; } public class GFG { // Function to insert a node at the beginning of the singly Linked List public static void Push( ref 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 to the new node head_ref = new_node; } // Function to find the digit sum for a number public static int DigitSum( int num) { int sum = 0; while (num != 0) { sum += (num % 10); num /= 10; } // Return the sum return sum; } // Recursive function to find the sum and product of nodes with even digit sum in a linked list public static void SumAndProductHelper(Node ptr, ref int prod, ref int sum) { // Base case: if the current node is null, return if (ptr == null ) { return ; } // Recursive call on the next node SumAndProductHelper(ptr.next, ref prod, ref sum); // If the current node has even digit sum and update the product and sum if ((DigitSum(ptr.data) & 1) == 0) { prod *= ptr.data; sum += ptr.data; } } // Wrapper function to call the recursive function public static void SumAndProduct(Node head_ref) { // Initialize the sum and product to 0 and 1 respectively int prod = 1; int sum = 0; // Call the recursive function on the head node SumAndProductHelper(head_ref, ref prod, ref sum); // Print the final Sum and Product Console.WriteLine( "Sum = " + sum); Console.WriteLine( "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 Push( ref head, 13); Push( ref head, 6); Push( ref head, 8); Push( ref head, 16); Push( ref head, 15); // Function call SumAndProduct(head); } } |
Javascript
// Node of Linked List class Node { constructor(data) { this .data = data; this .next = null ; } } // Function to insert a node at the // beginning of the singly Linked List function push(head_ref, new_data) { // Allocate new node let new_node = new Node(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 function digitSum(num) { let sum = 0; while (num) { sum += (num % 10); num = Math.floor(num / 10); } // Return the sum return sum; } // Recursive function to find the sum and product // of nodes with even digit sum in a linked list function sumAndProductHelper(ptr, prod, sum) { // Base case: if the current node is null, return if (ptr == null ) { return ; } // Recursive call on the next node sumAndProductHelper(ptr.next, prod, sum); // If the current node has even digit sum, // update the product and sum if (!(digitSum(ptr.data) & 1)) { prod[0] *= ptr.data; sum[0] += ptr.data; } } // Wrapper function to call the recursive function function sumAndProduct(head_ref) { // Initialise the sum and product to 0 and 1 respectively let prod = [1]; let sum = [0]; // Call the recursive function on the head node sumAndProductHelper(head_ref, prod, sum); // Print the final Sum and Product console.log( "Sum = " + sum[0]); console.log( "Product = " + prod[0]); } // Driver Code // Head of the linked list let 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); |
Sum = 42 Product = 9360
Time Complexity: O(n), where n is the number of nodes in the linked list.
Space Complexity: O(n), where n is the number of nodes in the linked list. This is because for each recursive call.