Find the product of last N nodes of the given Linked List
Last Updated :
06 Sep, 2022
Given a linked list and a number N. Find the product of last n nodes of the linked list.
Constraints : 0 <= N <= number of nodes in the linked list.
Examples:
Input : List = 10->6->8->4->12, N = 2
Output : 48
Explanation : Product of last two nodes:
12 * 4 = 48
Input : List = 15->7->9->5->16->14, N = 4
Output : 10080
Explanation : Product of last four nodes:
9 * 5 * 16 * 14 = 10080
Method 1:(Iterative approach using user-defined stack)
Traverse the nodes from left to right. While traversing push the nodes to a user-defined stack. Then pops the top n values from the stack and find their product.
Below is the implementation of the above approach:
C++
// C++ implementation to find the product of last
// 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
/* 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;
}
// utility function to find the product of last 'n' nodes
int productOfLastN_NodesUtil(struct Node* head, int n)
{
// if n == 0
if (n <= 0)
return 0;
stack<int> st;
int prod = 1;
// traverses the list from left to right
while (head != NULL) {
// push the node's data onto the stack 'st'
st.push(head->data);
// move to next node
head = head->next;
}
// pop 'n' nodes from 'st' and
// add them
while (n--) {
prod *= st.top();
st.pop();
}
// required product
return prod;
}
// Driver program to test above
int main()
{
struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << productOfLastN_NodesUtil(head, n);
return 0;
}
// Java implementation to find the product
// of last 'n' nodes of the Linked List
import java.util.*;
class GFG
{
/* A Linked list node */
static class Node
{
int data;
Node next;
};
static Node head;
// function to insert a node at the
// beginning of the linked list
static void 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;
head = head_ref;
}
// utility function to find the product
// of last 'n' nodes
static int productOfLastN_NodesUtil(Node head,
int n)
{
// if n == 0
if (n <= 0)
return 0;
Stack<Integer> st = new Stack<Integer>();
int prod = 1;
// traverses the list from left to right
while (head != null)
{
// push the node's data
// onto the stack 'st'
st.push(head.data);
// move to next node
head = head.next;
}
// pop 'n' nodes from 'st' and
// add them
while (n-- >0)
{
prod *= st.peek();
st.pop();
}
// required product
return prod;
}
// Driver Code
public static void main(String[] args)
{
head = null;
// create linked list 10->6->8->4->12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
int n = 2;
System.out.println(productOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by PrinciRaj1992
# Python3 implementation to find the product
# of last 'n' nodes of the Linked List
# Link list node
class Node:
def __init__(self, data):
self.data = data
self.next = next
head = None
# function to insert a node at the
# beginning of the linked list
def push(head_ref, new_data):
global head
# allocate node
new_node = Node(0)
# 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
head = head_ref
# utility function to find the product
# of last 'n' nodes
def productOfLastN_NodesUtil(head, n):
# if n == 0
if (n <= 0):
return 0
st = []
prod = 1
# traverses the list from left to right
while (head != None) :
# push the node's data
# onto the stack 'st'
st.append(head.data)
# move to next node
head = head.next
# pop 'n' nodes from 'st' and
# add them
while (n > 0) :
n = n - 1
prod *= st[-1]
st.pop()
# required product
return prod
# Driver Code
head = None
# create linked list 10->6->8->4->12
push(head, 12)
push(head, 4)
push(head, 8)
push(head, 6)
push(head, 10)
n = 2
print(productOfLastN_NodesUtil(head, n))
# This code is contributed by Arnab Kundu
// C# implementation to find the product
// of last 'n' nodes of the Linked List
using System;
class GFG
{
/* A Linked list node */
public 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;
}
// utility function to find the product
// of last 'n' nodes
static int productOfLastN_NodesUtil(Node head,
int n)
{
// if n == 0
if (n <= 0)
return 0;
int prod = 1, len = 0;
Node temp = head;
// calculate the length of the linked list
while (temp != null)
{
len++;
temp = temp.next;
}
// count of first (len - n) nodes
int c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != null && c-- >0)
// move to next node
temp = temp.next;
// now traverse the last 'n' nodes
// and add them
while (temp != null)
{
// accumulate node's data to sum
prod *= temp.data;
// move to next node
temp = temp.next;
}
// required product
return prod;
}
// Driver Code
public static void Main(String[] args)
{
Node head = null;
// create 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 n = 2;
Console.Write(productOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by Rajput-Ji
<script>
// javascript implementation to find the product
// of last 'n' nodes of the Linked List
/* A Linked list node */
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
var head;
// 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 point to the new node */
(head_ref) = new_node;
head = head_ref;
}
// utility function to find the product
// of last 'n' nodes
function productOfLastN_NodesUtil(head , n) {
// if n == 0
if (n <= 0)
return 0;
var st = [];
var prod = 1;
// traverses the list from left to right
while (head != null) {
// push the node's data
// onto the stack 'st'
st.push(head.data);
// move to next node
head = head.next;
}
// pop 'n' nodes from 'st' and
// add them
while (n-- > 0) {
prod *= st.pop();
}
// required product
return prod;
}
// Driver Code
head = null;
// create linked list 10->6->8->4->12
push(head, 12);
push(head, 4);
push(head, 8);
push(head, 6);
push(head, 10);
var n = 2;
document.write(productOfLastN_NodesUtil(head, n));
// This code contributed by aashish1995
</script>
Time complexity : O(n)
Method 2: (Recursive approach using system call stack)
Recursively traverse the linked list up to the end. Now during the return from the function calls, multiply the last n nodes. The product can be accumulated in some variable passed by reference to the function or to some global variable.
Below is the implementation of the above approach:
C++
// C++ implementation to find the product of
// last 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
/* 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 recursively find the product of last
// 'n' nodes of the given linked list
void productOfLastN_Nodes(struct Node* head, int* n,
int* prod)
{
// if head = NULL
if (!head)
return;
// recursively traverse the remaining nodes
productOfLastN_Nodes(head->next, n, prod);
// if node count 'n' is greater than 0
if (*n > 0) {
// accumulate sum
*prod = *prod * head->data;
// reduce node count 'n' by 1
--*n;
}
}
// utility function to find the product of last 'n' nodes
int productOfLastN_NodesUtil(struct Node* head, int n)
{
// if n == 0
if (n <= 0)
return 0;
int prod = 1;
// find the sum of last 'n' nodes
productOfLastN_Nodes(head, &n, &prod);
// required product
return prod;
}
// Driver program to test above
int main()
{
struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << productOfLastN_NodesUtil(head, n);
return 0;
}
// Java implementation to find the product of
// last 'n' nodes of the Linked List
class GFG{
static int n, prod;
/* 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 recursively find the product of last
// 'n' nodes of the given linked list
static void productOfLastN_Nodes(Node head)
{
// if head = null
if (head==null)
return;
// recursively traverse the remaining nodes
productOfLastN_Nodes(head.next);
// if node count 'n' is greater than 0
if (n > 0) {
// accumulate sum
prod = prod * head.data;
// reduce node count 'n' by 1
--n;
}
}
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head)
{
// if n == 0
if (n <= 0)
return 0;
prod = 1;
// find the sum of last 'n' nodes
productOfLastN_Nodes(head);
// required product
return prod;
}
// Driver program to test above
public static void main(String[] args)
{
Node head = null;
// create 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);
n = 2;
System.out.print(productOfLastN_NodesUtil(head));
}
}
//This code is contributed by 29AjayKumar
# Python implementation to find the product of
# last 'n' Nodes of the Linked List
n, prod = 0, 0;
''' A Linked list Node '''
class Node:
def __init__(self, data):
self.data = data
self.next = next
# function to insert a Node at the
# beginning of the linked list
def push(head_ref, new_data):
''' allocate Node '''
new_Node = Node(0);
''' 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 recursively find the product of last
# 'n' Nodes of the given linked list
def productOfLastN_Nodes(head):
global n, prod;
# if head = None
if (head == None):
return;
# recursively traverse the remaining Nodes
productOfLastN_Nodes(head.next);
# if Node count 'n' is greater than 0
if (n > 0):
# accumulate sum
prod = prod * head.data;
# reduce Node count 'n' by 1
n -= 1;
# utility function to find the product of last 'n' Nodes
def productOfLastN_NodesUtil(head):
global n,prod;
# if n == 0
if (n <= 0):
return 0;
prod = 1;
# find the sum of last 'n' Nodes
productOfLastN_Nodes(head);
# required product
return prod;
# Driver program to test above
if __name__ == '__main__':
head = None;
n = 2;
# create 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);
print(productOfLastN_NodesUtil(head));
# This code is contributed by 29AjayKumar
// C# implementation to find the product of
// last 'n' nodes of the Linked List
using System;
public class GFG{
static int n, prod;
/* A Linked list node */
public 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 recursively find the product of last
// 'n' nodes of the given linked list
static void productOfLastN_Nodes(Node head)
{
// if head = null
if (head==null)
return;
// recursively traverse the remaining nodes
productOfLastN_Nodes(head.next);
// if node count 'n' is greater than 0
if (n > 0) {
// accumulate sum
prod = prod * head.data;
// reduce node count 'n' by 1
--n;
}
}
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head)
{
// if n == 0
if (n <= 0)
return 0;
prod = 1;
// find the sum of last 'n' nodes
productOfLastN_Nodes(head);
// required product
return prod;
}
// Driver program to test above
public static void Main(String[] args)
{
Node head = null;
// create 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);
n = 2;
Console.Write(productOfLastN_NodesUtil(head));
}
}
// This code is contributed by 29AjayKumar
<script>
// JavaScript implementation to find the product of
// last 'n' nodes of the Linked List
var n, prod;
/* 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 point to the new node */
head_ref = new_node;
return head_ref;
}
// Function to recursively find the product of last
// 'n' nodes of the given linked list
function productOfLastN_Nodes(head) {
// if head = null
if (head == null)
return;
// recursively traverse the remaining nodes
productOfLastN_Nodes(head.next);
// if node count 'n' is greater than 0
if (n > 0) {
// accumulate sum
prod = prod * head.data;
// reduce node count 'n' by 1
--n;
}
}
// utility function to find the product of last 'n' nodes
function productOfLastN_NodesUtil(head) {
// if n == 0
if (n <= 0)
return 0;
prod = 1;
// find the sum of last 'n' nodes
productOfLastN_Nodes(head);
// required product
return prod;
}
// Driver program to test above
var head = null;
// create 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);
n = 2;
document.write(productOfLastN_NodesUtil(head));
// This code contributed by gauravrajput1
</script>
Time complexity: O(n)
Method 3 (Reversing the linked list):
Following are the steps:
- Reverse the given linked list.
- Traverse the first n nodes of the reversed linked list.
- While traversing multiply them.
- Reverse the linked list back to its original order.
- Return the product.
Below is the implementation of the above approach:
C++
// C++ implementation to find the product of last
// 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
/* 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;
}
void reverseList(struct Node** head_ref)
{
struct Node *current, *prev, *next;
current = *head_ref;
prev = NULL;
while (current != NULL) {
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
// utility function to find the product of last 'n' nodes
int productOfLastN_NodesUtil(struct Node* head, int n)
{
// if n == 0
if (n <= 0)
return 0;
// reverse the linked list
reverseList(&head);
int prod = 1;
struct Node* current = head;
// traverse the 1st 'n' nodes of the reversed
// linked list and product them
while (current != NULL && n--) {
// accumulate node's data to 'sum'
prod *= current->data;
// move to next node
current = current->next;
}
// reverse back the linked list
reverseList(&head);
// required product
return prod;
}
// Driver program to test above
int main()
{
struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << productOfLastN_NodesUtil(head, n);
return 0;
}
// Java implementation to find the product of last
// 'n' nodes of the Linked List
class GFG
{
/* 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;
}
static Node reverseList(Node head_ref)
{
Node current, prev, next;
current = head_ref;
prev = null;
while (current != null)
{
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head_ref = prev;
return head_ref;
}
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head, int n)
{
// if n == 0
if (n <= 0)
return 0;
// reverse the linked list
head = reverseList(head);
int prod = 1;
Node current = head;
// traverse the 1st 'n' nodes of the reversed
// linked list and product them
while (current != null && n-- >0)
{
// accumulate node's data to 'sum'
prod *= current.data;
// move to next node
current = current.next;
}
// reverse back the linked list
head = reverseList(head);
// required product
return prod;
}
// Driver program to test above
public static void main(String[] args)
{
Node head = null;
// create 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 n = 2;
System.out.print(productOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by Rajput-Ji
# Python3 implementation to find the product of last
# 'n' nodes of the Linked List
''' 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 point to the new node '''
(head_ref) = new_node;
return head_ref
def reverseList(head_ref):
next = None
current = head_ref;
prev = None;
while (current != None):
next = current.next;
current.next = prev;
prev = current;
current = next;
head_ref = prev
return head_ref
# utility function to find the product of last 'n' nodes
def productOfLastN_NodesUtil(head, n):
# if n == 0
if (n <= 0):
return 0;
# reverse the linked list
head = reverseList(head);
prod = 1;
current = head;
# traverse the 1st 'n' nodes of the reversed
# linked list and product them
while (current != None and n):
n -= 1
# accumulate node's data to 'sum'
prod *= current.data;
# move to next node
current = current.next;
# reverse back the linked list
head = reverseList(head);
# required product
return prod;
# Driver program to test above
if __name__=='__main__':
head = None;
# create 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);
n = 2;
print(productOfLastN_NodesUtil(head, n))
# This code is contributed by rutvik_56
// C# implementation to find
// the product of last 'n' 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;
}
static Node reverseList(Node head_ref)
{
Node current, prev, next;
current = head_ref;
prev = null;
while (current != null)
{
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head_ref = prev;
return head_ref;
}
// utility function to find
// the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head, int n)
{
// if n == 0
if (n <= 0)
return 0;
// reverse the linked list
head = reverseList(head);
int prod = 1;
Node current = head;
// traverse the 1st 'n' nodes of the reversed
// linked list and product them
while (current != null && n-- >0)
{
// accumulate node's data to 'sum'
prod *= current.data;
// move to next node
current = current.next;
}
// reverse back the linked list
head = reverseList(head);
// required product
return prod;
}
// Driver Code
public static void Main(String[] args)
{
Node head = null;
// create 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 n = 2;
Console.Write(productOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by 29AjayKumar
<script>
// javascript implementation to find the product of last
// 'n' 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 point to the new node */
head_ref = new_node;
return head_ref;
}
function reverseList(head_ref) {
var current, prev, next;
current = head_ref;
prev = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head_ref = prev;
return head_ref;
}
// utility function to find the product of last 'n' nodes
function productOfLastN_NodesUtil(head , n) {
// if n == 0
if (n <= 0)
return 0;
// reverse the linked list
head = reverseList(head);
var prod = 1;
var current = head;
// traverse the 1st 'n' nodes of the reversed
// linked list and product them
while (current != null && n-- > 0) {
// accumulate node's data to 'sum'
prod *= current.data;
// move to next node
current = current.next;
}
// reverse back the linked list
head = reverseList(head);
// required product
return prod;
}
// Driver program to test above
var head = null;
// create 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 n = 2;
document.write(productOfLastN_NodesUtil(head, n));
// This code contributed by aashish1995
</script>
Time complexity: O(n)
Method 4 (Using a length of linked list):
Following are the steps:
- Calculate the length of the given Linked List. Let it be len.
- First traverse the (len – n) nodes from the beginning.
- Then traverse the remaining n nodes and while traversing product them.
- Return the product.
Below is the implementation of the above approach:
C++
// C++ implementation to find the product of last
// 'n' nodes of the Linked List
#include <bits/stdc++.h>
using namespace std;
/* 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;
}
// utility function to find the product of last 'n' nodes
int productOfLastN_NodesUtil(struct Node* head, int n)
{
// if n == 0
if (n <= 0)
return 0;
int prod = 1, len = 0;
struct Node* temp = head;
// calculate the length of the linked list
while (temp != NULL) {
len++;
temp = temp->next;
}
// count of first (len - n) nodes
int c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != NULL && c--)
// move to next node
temp = temp->next;
// now traverse the last 'n' nodes and add them
while (temp != NULL) {
// accumulate node's data to sum
prod *= temp->data;
// move to next node
temp = temp->next;
}
// required product
return prod;
}
// Driver program to test above
int main()
{
struct Node* head = NULL;
// create linked list 10->6->8->4->12
push(&head, 12);
push(&head, 4);
push(&head, 8);
push(&head, 6);
push(&head, 10);
int n = 2;
cout << productOfLastN_NodesUtil(head, n);
return 0;
}
// Java implementation to find the product of last
// 'n' nodes of the Linked List
class GFG
{
/* 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;
}
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head, int n)
{
// if n == 0
if (n <= 0)
return 0;
int prod = 1, len = 0;
Node temp = head;
// calculate the length of the linked list
while (temp != null)
{
len++;
temp = temp.next;
}
// count of first (len - n) nodes
int c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != null && c-- >0)
// move to next node
temp = temp.next;
// now traverse the last 'n' nodes and add them
while (temp != null)
{
// accumulate node's data to sum
prod *= temp.data;
// move to next node
temp = temp.next;
}
// required product
return prod;
}
// Driver program to test above
public static void main(String[] args)
{
Node head = null;
// create 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 n = 2;
System.out.print(productOfLastN_NodesUtil(head, n));
}
}
// This code is contributed by PrinciRaj1992
# Python3 implementation to find the product of last
# 'n' nodes of the Linked List
''' A Linked list node '''
class Node:
def __init__(self):
self.data = 0
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();
''' 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;
# utility function to find the product of last 'n' nodes
def productOfLastN_NodesUtil(head, n):
# if n == 0
if (n <= 0):
return 0;
prod = 1
len = 0;
temp = head;
# calculate the length of the linked list
while (temp != None):
len += 1
temp = temp.next;
# count of first (len - n) nodes
c = len - n;
temp = head;
# just traverse the 1st 'c' nodes
while (temp != None and c > 0):
c -= 1
# move to next node
temp = temp.next;
# now traverse the last 'n' nodes and add them
while (temp != None):
# accumulate node's data to sum
prod *= temp.data;
# move to next node
temp = temp.next;
# required product
return prod;
# Driver program to test above
if __name__=='__main__':
head = None;
# create 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);
n = 2;
print(productOfLastN_NodesUtil(head, n));
# This code is contributed by Pratham76
// C# implementation to find the product of last
// 'n' nodes of the Linked List
using System;
public class GFG
{
/* A Linked list node */
public 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;
}
// utility function to find the product of last 'n' nodes
static int productOfLastN_NodesUtil(Node head, int n)
{
// if n == 0
if (n <= 0)
return 0;
int prod = 1, len = 0;
Node temp = head;
// calculate the length of the linked list
while (temp != null)
{
len++;
temp = temp.next;
}
// count of first (len - n) nodes
int c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != null && c-- >0)
// move to next node
temp = temp.next;
// now traverse the last 'n' nodes and add them
while (temp != null)
{
// accumulate node's data to sum
prod *= temp.data;
// move to next node
temp = temp.next;
}
// required product
return prod;
}
// Driver program to test above
public static void Main(String[] args)
{
Node head = null;
// create 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 n = 2;
Console.Write(productOfLastN_NodesUtil(head, n));
}
}
// This code contributed by Rajput-Ji
<script>
// Javascript implementation to find the product of last
// 'n' nodes of the Linked List
/* A Linked list node */
class Node
{
constructor()
{
this.data = 0;
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 point to the new node */
head_ref = new_node;
return head_ref;
}
// utility function to find the product of last 'n' nodes
function productOfLastN_NodesUtil(head, n)
{
// if n == 0
if (n <= 0)
return 0;
var prod = 1, len = 0;
var temp = head;
// calculate the length of the linked list
while (temp != null)
{
len++;
temp = temp.next;
}
// count of first (len - n) nodes
var c = len - n;
temp = head;
// just traverse the 1st 'c' nodes
while (temp != null && c-- >0)
// move to next node
temp = temp.next;
// now traverse the last 'n' nodes and add them
while (temp != null)
{
// accumulate node's data to sum
prod *= temp.data;
// move to next node
temp = temp.next;
}
// required product
return prod;
}
// Driver program to test above
var head = null;
// create 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 n = 2;
document.write(productOfLastN_NodesUtil(head, n));
</script>
Time complexity: O(n)
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