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g2801_2900.s2861_maximum_number_of_alloys.readme.md Maven / Gradle / Ivy

2861\. Maximum Number of Alloys

Medium

You are the owner of a company that creates alloys using various types of metals. There are `n` different types of metals available, and you have access to `k` machines that can be used to create alloys. Each machine requires a specific amount of each metal type to create an alloy.

For the ith machine to create an alloy, it needs `composition[i][j]` units of metal of type `j`. Initially, you have `stock[i]` units of metal type `i`, and purchasing one unit of metal type `i` costs `cost[i]` coins.

Given integers `n`, `k`, `budget`, a **1-indexed** 2D array `composition`, and **1-indexed** arrays `stock` and `cost`, your goal is to **maximize** the number of alloys the company can create while staying within the budget of `budget` coins.

**All alloys must be created with the same machine.**

Return _the maximum number of alloys that the company can create_.

**Example 1:**

**Input:** n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]

**Output:** 2

**Explanation:** It is optimal to use the 1st machine to create alloys. 

To create 2 alloys we need to buy the: 
- 2 units of metal of the 1st type. 
- 2 units of metal of the 2nd type. 
- 2 units of metal of the 3rd type. 

In total, we need 2 \* 1 + 2 \* 2 + 2 \* 3 = 12 coins, which is smaller than or equal to budget = 15. Notice that we have 0 units of metal of each type and we have to buy all the required units of metal. It can be proven that we can create at most 2 alloys.

**Example 2:**

**Input:** n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,100], cost = [1,2,3]

**Output:** 5

**Explanation:** It is optimal to use the 2nd machine to create alloys. To create 5 alloys we need to buy: 
- 5 units of metal of the 1st type. 
- 5 units of metal of the 2nd type. 
- 0 units of metal of the 3rd type. 

In total, we need 5 \* 1 + 5 \* 2 + 0 \* 3 = 15 coins, which is smaller than or equal to budget = 15. It can be proven that we can create at most 5 alloys.

**Example 3:**

**Input:** n = 2, k = 3, budget = 10, composition = [[2,1],[1,2],[1,1]], stock = [1,1], cost = [5,5]

**Output:** 2

**Explanation:** It is optimal to use the 3rd machine to create alloys. To create 2 alloys we need to buy the: 
- 1 unit of metal of the 1st type. 
- 1 unit of metal of the 2nd type. 

In total, we need 1 \* 5 + 1 \* 5 = 10 coins, which is smaller than or equal to budget = 10. It can be proven that we can create at most 2 alloys.

**Constraints:**

*   `1 <= n, k <= 100`
*   0 <= budget <= 108
*   `composition.length == k`
*   `composition[i].length == n`
*   `1 <= composition[i][j] <= 100`
*   `stock.length == cost.length == n`
*   0 <= stock[i] <= 108
*   `1 <= cost[i] <= 100`