# Find the product of last N nodes of the given Linked List in C++

C++Server Side ProgrammingProgramming

Consider we have few elements in a linked list. We have to find the multiplication result of last n number of elements. The value of n is also given. So if the list is like [5, 7, 3, 5, 6, 9], and n = 3, then result will be 5 * 6 * 9 = 270.

The process is straight forward. We simply read the current element starting from left side, then add the elements into stack. After filling up the stack, remove n elements and multiply them with the prod. (initially prod is 1), when n number of elements are traversed, then stop.

## Example

#include<iostream>
#include<stack>
using namespace std;
class Node{
public:
int data;
Node *next;
};
Node* getNode(int data){
Node *newNode = new Node;
newNode->data = data;
newNode->next = NULL;
return newNode;
}
void append(struct Node** start, int key) {
Node* new_node = getNode(key);
Node *p = (*start);
if(p == NULL){
(*start) = new_node;
return;
}
while(p->next != NULL){
p = p->next;
}
p->next = new_node;
}
long long prodLastNElements(Node *start, int n) {
if(n <= 0)
return 0;
stack<int> stk;
long long res = 1;
Node* temp = start;
while (temp != NULL) {
stk.push(temp->data);
temp = temp->next;
}
while(n--){
res *= stk.top();
stk.pop();
}
return res;
}
int main() {
Node *start = NULL;
int arr[] = {5, 7, 3, 5, 6, 9};
int size = sizeof(arr)/sizeof(arr[0]);
int n = 3;
for(int i = 0; i<size; i++){
append(&start, arr[i]);
}
cout << "Product of last n elements: " << prodLastNElements(start, n);
}

## Output

Product of last n elements: 270
Published on 04-Nov-2019 14:29:11