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 an amount of **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, 2 stocks on day 2 and 1 stock on day 3 for 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 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, substract 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 C++ implementation of this approach:

`// CPP 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; ` `} ` |

Output:

4

**Time Complexity :**O(nlogn).

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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