Count All Sub sequences having Product less than K - c++

Question

Given a non-negative array, find the number of subsequences having a product smaller than K.
Examples:
Input : [1, 2, 3, 4]
k = 10
Output :11
Input : [4, 8, 7, 2]
k = 50
Output : 9
So, We want to count the number of subsequences whose product is less than K.
There are sub-problems, and it can be solved using Dynamic Programming
However, I tried to write down the recursive code for better understanding.
Note: I am getting an answer as 6, which is wrong.
Can someone help me, How to foresee the correct Logic?
#include <bits/stdc++.h>
using namespace std;
vector<int> A{1, 2, 3, 4};
int countSubsequence(int i, int prod, int K)
{
if(prod > 1 && prod <= K)
return 1;
if(i >= A.size() || prod > K)
return 0;
return countSubsequence(i + 1, prod, K) + countSubsequence(i + 1, prod*A[i], K);
}
int main()
{
int K = 10;
cout << countSubsequence(0, 1, K);
return 0;
}

Answer 1

The condition
if(prod > 1 && prod <= K)
return 1;
will have it return from the function (for example) when [1, 2] is selected from [1, 2, 3, 4] and prevent it from searching for [1, 2, 3].
Also:
The condition prod <= K is wrong becasue you want the product smaller than K, not K or smaller.
You cannot distinguish "nothing is multiplied" and "only the number 1 is multiplied" when you use 1 as the initial value.
Try this:
#include <bits/stdc++.h>
using namespace std;
vector<int> A{1, 2, 3, 4};
int countSubsequence(int i, int prod, int K)
{
if(i >= A.size() && 0 <= prod && prod < K)
return 1;
if(i >= A.size() || prod >= K)
return 0;
return countSubsequence(i + 1, prod, K) + countSubsequence(i + 1, prod < 0 ? A[i] : prod*A[i], K);
}
int main()
{
int K = 10;
cout << countSubsequence(0, -1, K);
return 0;
}