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Find the product of last N nodes of the given Linked List

  • Last Updated : 11 Nov, 2021

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

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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




// 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




# 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#




// 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

Javascript




<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>
Output:
48

 

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




// 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

Python3




# 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 poto 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#




// 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

Javascript




<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>
Output:
48

 

Time complexity: O(n)
 

Method 3 (Reversing the linked list)

Following are the steps:

  1. Reverse the given linked list.
  2. Traverse the first n nodes of the reversed linked list.
  3. While traversing multiply them.
  4. Reverse the linked list back to its original order.
  5. 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




// 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




# 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#




// 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

Javascript




<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>
Output:
48

 

Time complexity: O(n)
 

Method 4 (Using length of linked list)

Following are the steps:

  1. Calculate the length of the given Linked List. Let it be len.
  2. First traverse the (len – n) nodes from the beginning.
  3. Then traverse the remaining n nodes and while traversing product them.
  4. 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




// 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




# 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#




// 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

Javascript




<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>
Output:
48

 

Time complexity : O(n)
 




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