# Product of the Last K Numbers in C++

C++Server Side ProgrammingProgramming

Suppose we have the implement the class called ProductOfNumbers that supports two methods −

• add(int num): This adds the number num to the back of the current list of numbers.

• getProduct(int k): This returns the product of the last k numbers in the current list.

We can assume that always the current list has at least k numbers. So for example, if the input is like − add(3), add(0), add(2), add(5), add(4), getProduct(2), getProduct(3), getProduct(4), add(8), getProduct(2), then the output will be (after each function call) −

[3], [3, 0], [3, 0, 2], [3, 0, 2, 5], [3, 0, 2, 5, 4], then (5 * 4) = 20, then (2 * 5 * 4) = 40, then (0 * 2 * 5 * 4) = 0, then [3, 0, 2, 5, 4, 8], then (4 * 8) = 32.

To solve this, we will follow these steps −

• In the initialization section, it will create an array, and put 1 into it

• The add() method will take num

• if num is 0, then clear the array, and insert 1, otherwise insert last_element * num into array

• The getProduct() method will take k as input

• n := size of the array

• if k > n – 1, then return 0, otherwise dp[n - 1] / dp[n – k – 1]

## Example (C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class ProductOfNumbers {
public:
vector <int> dq;
ProductOfNumbers() {
dq.push_back(1);
}
void add(int num) {
if(num == 0){
dq.clear();
dq.push_back(1);
}
else{
dq.push_back(dq.back() * num);
}
}
int getProduct(int k) {
int n = (int)dq.size();
return k > n - 1? 0 : dq[n - 1] / dq[n - k - 1];
}
};
main(){
ProductOfNumbers ob;
cout << (ob.getProduct(2)) << endl;
cout << (ob.getProduct(3)) << endl;
cout << (ob.getProduct(4)) << endl;
cout << (ob.getProduct(2)) << endl;
}

## Input

add(3)
getProduct(2)
getProduct(3)
getProduct(4)
getProduct(2)
20
32