Find the product of first k nodes of the given Linked List
Given a pointer to the head of a singly linked list and an integer k. The task is to find the product of first k nodes of the linked list.
Examples:
Input: 10 -> 6 -> 8 -> 4 -> 12, k = 2
Output: 60
10 * 6 = 60Input: 15 -> 7 -> 9 -> 5 -> 16 -> 14, k = 4
Output: 4725
15 * 7 * 9 * 5 = 4725
Approach: Set prod = 1 (required product) and count = 0 (count of nodes traversed). Now, start traversing the nodes of the linked list from left to right and update count = count + 1 and prod = prod * currNode -> data with every traversed node while count < k. Print the value of prod in the end.
Below is the implementation of the above approach:
C++
// C++ program to find the product of first // 'k' nodes of the Linked List #include <bits/stdc++.h> using namespace std; #define ll long long int /* 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 return the product of // first k nodes of the given linked list ll product(struct Node* head, int k) { if (k <= 0) return 0; ll prod = 1; int i = 0; Node* node = head; // Traverse the list from left to right while (i < k) { // Update product prod = prod * node->data; // Move to the next node node = node->next; i++; } // Return the required product return prod; } // Driver code int main() { struct Node* head = NULL; // Linked list 10 -> 6 -> 8 -> 4 -> 12 push(&head, 12); push(&head, 4); push(&head, 8); push(&head, 6); push(&head, 10); int k = 2; cout << product(head, k); return 0; } |
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Java
// Java program to find the product of first // 'k' nodes of the Linked List class Solution { /* 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 return the product of // first k nodes of the given linked list static long product( Node head, int k) { if (k <= 0) return 0; long prod = 1; int i = 0; Node node = head; // Traverse the list from left to right while (i < k) { // Update product prod = prod * node.data; // Move to the next node node = node.next; i++; } // Return the required product return prod; } // Driver code public static void main(String args[]) { Node head = new Node(); // Linked list 10 . 6 . 8 . 4 . 12 head=push(head, 12); head=push(head, 4); head=push(head, 8); head=push(head, 6); head=push(head, 10); int k = 2; System.out.println( product(head, k)); } } // This code is contributed by Arnab Kundu |
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Python3
# Python3 program to find the product # of first 'k' nodes of the Linked List import math #define ll long long # 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 po to the new node head_ref = new_node return head_ref # Function to return the product of # first k nodes of the given linked list def product(head, k): if (k <= 0): return 0 prod = 1 i = 0 node = head # Traverse the list from left to right while (i < k): # Update product prod = prod * node.data # Move to the next node node = node.next i = i + 1 # Return the required product return prod # Driver code if __name__=='__main__': head = None # Linked list 10 . 6 . 8 . 4 . 12 head = push(head, 12); head = push(head, 4); head = push(head, 8); head = push(head, 6); head = push(head, 10); k = 2 print(product(head, k)) # This code is contributed by Srathore |
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C#
// C# program to find the product of first // 'k' nodes of the Linked List using System; class GFG { /* A Linked list node */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 return the product of // first k nodes of the given linked list static long product( Node head, int k) { if (k <= 0) return 0; long prod = 1; int i = 0; Node node = head; // Traverse the list from left to right while (i < k) { // Update product prod = prod * node.data; // Move to the next node node = node.next; i++; } // Return the required product return prod; } // Driver code public static void Main() { Node head = new Node(); // Linked list 10 . 6 . 8 . 4 . 12 head=push(head, 12); head=push(head, 4); head=push(head, 8); head=push(head, 6); head=push(head, 10); int k = 2; Console.WriteLine( product(head, k)); } } // This code is contributed by PrinciRaj1992 |
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Output:
60
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