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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 st = ``new` `Stack(); ` `    ``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  `

## C#

 `// C# implementation to find the product  ` `// of last 'n' nodes of the Linked List ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG  ` `{ ` ` `  `/* A Linked list node */` `public` `class` `Node  ` `{ ` `    ``public` `int` `data; ` `    ``public` `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<``int``> st = ``new` `Stack<``int``>(); ` `    ``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; ` `    ``Console.WriteLine(productOfLastN_NodesUtil(head, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

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++ implementation to find the product of ` `// last 'n' nodes of the Linked List ` `#include ` `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; ` `} `

Output:

```48
```

Time complexity : O(n)
Method 3 (Reversing the linked list):

1. Following are the steps:
2. Reverse the given linked list.
3. Traverse the first n nodes of the reversed linked list.
4. While traversing multiply them.
5. Reverse the linked list back to its original order.
6. Return the product.

Below is the implementation of the above approach:

 `// C++ implementation to find the product of last ` `// 'n' nodes of the Linked List ` `#include ` `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; ` `} `

Output:

```48
```

Time complexity : O(n)
Method 4 (Using length of linked list):

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

Below is the implementation of the above approach:

 `// C++ implementation to find the product of last ` `// 'n' nodes of the Linked List ` `#include ` `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; ` `} `

Output:

```48
```

Time complexity : O(n)

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