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

<p>For the <code>i<sup>th</sup></code> machine to create an alloy, it needs <code>composition[i][j]</code> units of metal of type <code>j</code>. Initially, you have <code>stock[i]</code> units of metal type <code>i</code>, and purchasing one unit of metal type <code>i</code> costs <code>cost[i]</code> coins.</p>

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

<p><strong>All alloys must be created with the same machine.</strong></p>

<p>Return <em>the maximum number of alloys that the company can create</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,0], cost = [1,2,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> It is optimal to use the 1<sup>st</sup> machine to create alloys.
To create 2 alloys we need to buy the:
- 2 units of metal of the 1<sup>st</sup> type.
- 2 units of metal of the 2<sup>nd</sup> type.
- 2 units of metal of the 3<sup>rd</sup> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, k = 2, budget = 15, composition = [[1,1,1],[1,1,10]], stock = [0,0,100], cost = [1,2,3]
<strong>Output:</strong> 5
<strong>Explanation:</strong> It is optimal to use the 2<sup>nd</sup> machine to create alloys.
To create 5 alloys we need to buy:
- 5 units of metal of the 1<sup>st</sup> type.
- 5 units of metal of the 2<sup>nd</sup> type.
- 0 units of metal of the 3<sup>rd</sup> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 2, k = 3, budget = 10, composition = [[2,1],[1,2],[1,1]], stock = [1,1], cost = [5,5]
<strong>Output:</strong> 2
<strong>Explanation:</strong> It is optimal to use the 3<sup>rd</sup> machine to create alloys.
To create 2 alloys we need to buy the:
- 1 unit of metal of the 1<sup>st</sup> type.
- 1 unit of metal of the 2<sup>nd</sup> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n, k &lt;= 100</code></li>
	<li><code>0 &lt;= budget &lt;= 10<sup>8</sup></code></li>
	<li><code>composition.length == k</code></li>
	<li><code>composition[i].length == n</code></li>
	<li><code>1 &lt;= composition[i][j] &lt;= 100</code></li>
	<li><code>stock.length == cost.length == n</code></li>
	<li><code>0 &lt;= stock[i] &lt;= 10<sup>8</sup></code></li>
	<li><code>1 &lt;= cost[i] &lt;= 100</code></li>
</ul>
Array; Binary Search
func maxNumberOfAlloys(n int, k int, budget int, composition [][]int, stock []int, cost []int) int {
	isValid := func(target int) bool {
		for _, currMachine := range composition {
			remain := budget
			for i, x := range currMachine {
				need := max(0, x*target-stock[i])
				remain -= need * cost[i]
			}
			if remain >= 0 {
				return true
			}
		}
		return false
	}

	l, r := 0, budget+stock[0]
	for l < r {
		mid := (l + r + 1) >> 1
		if isValid(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	return l
}