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Product of the Last K Numbers in C++
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 −
#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;
(ob.add(3));
(ob.add(0));
(ob.add(2));
(ob.add(5));
(ob.add(4));
cout << (ob.getProduct(2)) << endl;
cout << (ob.getProduct(3)) << endl;
cout << (ob.getProduct(4)) << endl;
(ob.add(8));
cout << (ob.getProduct(2)) << endl;
}Input
add(3) add(0) add(2) add(5) add(4) getProduct(2) getProduct(3) getProduct(4) add(8) getProduct(2)
Output
20 40 0 32
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