C++ code to find money after buying and selling shares

Suppose we have two arrays A of size n, and B of size m, and another number r. There are n opportunities to buy shares. The i-th of them allows to buy as many shares as we want, ith share price is A[i]. And also there are m opportunities to sell shares. The i-th of them allows to sell as many shares as we want, sell price of ith share is B[i]. We can't sell more shares than we have. If we have r amount of money and no existing share, we have to find the maximum number of money we can hold after buying and selling.

So, if the input is like A = [4, 2, 5]; B = [4, 4, 5, 4]; r = 11, then the output will be 26, because we have 11 amount of money. It's optimal to buy 5 shares of a stock at the price of 2, and then to sell all of them at the price of 5. Thus we can get 26 at the end.

Steps

To solve this, we will follow these steps −

n := size of A
an := 1100
bn := 0
for initialize i := 0, when i < n, update (increase i by 1), do:
   if an > A[i], then:
      an := A[i]
for initialize i := 0, when i < m, update (increase i by 1), do:
   if bn < B[i], then:
      bn := B[i]
if bn > an, then:
   r := bn * (r / an) + (r - (r / an) * an)
return r

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(vector<int> A, vector<int> B, int r){
   int n = A.size(), m = B.size();
   int an = 1100, bn = 0;
   for (int i = 0; i < n; i++){
      if (an > A[i])
         an = A[i];
   }
   for (int i = 0; i < m; i++){
      if (bn < B[i])
         bn = B[i];
   }
   if (bn > an){
      r = (bn) * (r / an) + (r - (r / an) * an);
   }
   return r;
}
int main(){
   vector<int> A = { 4, 2, 5 };
   vector<int> B = { 4, 4, 5, 4 };
   int r = 11;
   cout << solve(A, B, r) << endl;
}