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 the 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 the 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 listvoid 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 listll 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 codeint 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;} |
Java
// Java program to find the product of first// 'k' nodes of the Linked Listclass Solution{/* A Linked list node */static class Node{ int data; Node next;}// Function to insert a node at the// beginning of the linked liststatic 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 liststatic 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 codepublic 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 |
Python3
# Python3 program to find the product# of first 'k' nodes of the Linked Listimport math#define ll long long# A Linked list nodeclass Node: def __init__(self, data): self.data = data self.next = None# Function to insert a node at the# beginning of the linked listdef 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 listdef 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 codeif __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 |
C#
// C# program to find the product of first// 'k' nodes of the Linked Listusing 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 liststatic 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 liststatic 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 codepublic 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 |
Javascript
<script>// javascript program to find the product of first// 'k' nodes of the Linked List/* A Linked list node */ class Node { constructor(val) { this.data = val; this.next = null; } } // Function to insert a node at the // beginning of the linked list function push(head_ref , new_data) { /* allocate node */ var 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 povar 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 function product(head , k) { if (k <= 0) return 0; var prod = 1; var i = 0; var 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 var 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); var k = 2; document.write(product(head, k));// This code contributed by Rajput-Ji</script> |
Output:
60
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