Buy Maximum Stocks if i stocks can be bought on i-th day

In a stock market, there is a product with its infinite stocks. The stock prices are given for N days, where arr[i] denotes the price of the stock on the i^th day. There is a rule that a customer can buy at most i stock on the i^th day. If the customer has k amount of money initially, find out the maximum number of stocks a customer can buy. 

For example, for 3 days the price of a stock is given as 7, 10, 4. You can buy 1 stock worth 7 rs on day 1, 2 stocks worth 10 rs each on day 2 and 3 stock worth 4 rs each on day 3.

Examples: 

Input : price[] = { 10, 7, 19 }, 
              k = 45.
Output : 4
A customer purchases 1 stock on day 1 for 10 rs, 
2 stocks on day 2 for 7 rs each and 1 stock on day 3 for 19 rs.Therefore total of
10, 7 * 2 = 14 and 19 respectively. Hence, 
total amount is 10 + 14 + 19 = 43 and number 
of stocks purchased is 4.

Input  : price[] = { 7, 10, 4 }, 
               k = 100.
Output : 6

The idea is to use greedy approach, where we should start buying product from the day when the stock price is least and so on. 
So, we will sort the pair of two values i.e { stock price, day } according to the stock price, and if stock prices are same, then we sort according to the day. 
Now, we will traverse along the sorted list of pairs, and start buying as follows: 
Let say, we have R rs remaining till now, and the cost of product on this day be C, and we can buy atmost L products on this day then, 
total purchase on this day (P) = min(L, R/C) 
Now, add this value to the answer 
total_purchase = total_purchase + P, where P =min(L, R/C) 
Now, subtract the cost of buying P items from remaining money, R = R – P*C. 
Total number of products that we can buy is equal to total_purchase.

Below is the implementation of this approach:  

C++

// C++ program to find maximum number of stocks that
// can be bought with given constraints.
#include <bits/stdc++.h>
using namespace std;

// Return the maximum stocks
int buyMaximumProducts(int n, int k, int price[])
{
    vector<pair<int, int> > v;

    // Making pair of product cost and number
    // of day..
    for (int i = 0; i < n; ++i)
        v.push_back(make_pair(price[i], i + 1));

    // Sorting the vector pair.
    sort(v.begin(), v.end());

    // Calculating the maximum number of stock
    // count.
    int ans = 0;
    for (int i = 0; i < n; ++i) {
        ans += min(v[i].second, k / v[i].first);
        k -= v[i].first * min(v[i].second,
                               (k / v[i].first));
    }

    return ans;
}

// Driven Program
int main()
{
    int price[] = { 10, 7, 19 };
    int n = sizeof(price)/sizeof(price[0]);
    int k = 45;

    cout << buyMaximumProducts(n, k, price) << endl;

    return 0;
}

Java

// Java program to find maximum number of stocks that
// can be bought with given constraints.
import java.util.*;

public class GFG {

    // Return the maximum stocks
    static int buyMaximumProducts(int[] price, int K, int n)
    {
        Pair[] arr = new Pair[n];

        // Making pair of product cost and number of day..
        for (int i = 0; i < n; i++)
            arr[i] = new Pair(price[i], i + 1);

        // Sorting the pair array.
        Arrays.sort(arr, new SortPair());
        // Calculating the maximum number of stock
        // count.
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans += Math.min(arr[i].second,
                            K / arr[i].first);
            K -= arr[i].first
                 * Math.min(arr[i].second,
                            K / arr[i].first);
        }
        return ans;
    }

  // Driver code
    public static void main(String[] args)
    {
        int[] price = { 10, 7, 19 };
        int K = 45;

        // int []price = { 7, 10, 4 };
        // int K = 100;
        System.out.println(
            buyMaximumProducts(price, K, price.length));
    }
}

// Helper class
class Pair {
    int first, second;
    Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// For Sorting using Pair.first value
class SortPair implements Comparator<Pair> {
    public int compare(Pair a, Pair b)
    {
        return a.first - b.first;
    }
}

// This code is contributed by Aakash Choudhary

Python3

# Python3 program to find maximum number of stocks
# that can be bought with given constraints.

# Returns the maximum stocks
def buyMaximumProducts(n, k, price):

    # Making pair of stock cost and day number
    arr = []

    for i in range(n):
        arr.append([i + 1, price[i]])

    # Sort based on the price of stock
    arr.sort(key = lambda x: x[1])

    # Calculating the max stocks purchased
    total_purchase = 0
    for i in range(n):
        P = min(arr[i][0], k//arr[i][1])
        total_purchase += P
        k -= (P * arr[i][1])

    return total_purchase

# Driver code
price = [ 10, 7, 19 ]
n = len(price)
k = 45

print(buyMaximumProducts(n, k, price))

# This code is contributed by Tharun Reddy

C#

// C# program to find maximum number of stocks that
// can be bought with given constraints.
using System;

class GFG
{

  // Driver code
  static void Main(string[] args)
  {

    int[] price = { 10, 7, 19 };
    int K = 45;

    // int []price = { 7, 10, 4 };
    // int K = 100;
    Console.WriteLine(
      buyMaximumProducts(price, K, price.Length));
  }

  // Return the maximum stocks
  static int buyMaximumProducts(int[] price, int K, int n)
  {

    Pair[] arr = new Pair[n];

    // Making pair of product cost and number of day..
    for (int i = 0; i < n; i++)
      arr[i] = new Pair(price[i], i + 1);

    // Sorting the pair array.
    Array.Sort(arr);

    // Calculating the maximum number of stock count.
    int ans = 0;
    for (int i = 0; i < n; i++) {

      ans += Math.Min(arr[i].second,
                      K / arr[i].first);

      K -= arr[i].first
        * Math.Min(arr[i].second,
                   K / arr[i].first);
    }

    return ans;
  }
}

// Helper class
class Pair : IComparable<Pair> {

  public int first, second;

  public Pair(int first, int second)
  {
    this.first = first;
    this.second = second;
  }

  // For Sorting using Pair.first value
  public int CompareTo(Pair other)
  {
    return this.first - other.first;
  }
}

// This code is contributed by Tapesh (tapeshdua420)

Output

4

Time Complexity: O(nlogn).
Auxiliary Space: O(n)

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