<p>There is a school that has classes of students and each class will be having a final exam. You are given a 2D integer array <code>classes</code>, where <code>classes[i] = [pass<sub>i</sub>, total<sub>i</sub>]</code>. You know beforehand that in the <code>i<sup>th</sup></code> class, there are <code>total<sub>i</sub></code> total students, but only <code>pass<sub>i</sub></code> number of students will pass the exam.</p>

<p>You are also given an integer <code>extraStudents</code>. There are another <code>extraStudents</code> brilliant students that are <strong>guaranteed</strong> to pass the exam of any class they are assigned to. You want to assign each of the <code>extraStudents</code> students to a class in a way that <strong>maximizes</strong> the <strong>average</strong> pass ratio across <strong>all</strong> the classes.</p>

<p>The <strong>pass ratio</strong> of a class is equal to the number of students of the class that will pass the exam divided by the total number of students of the class. The <strong>average pass ratio</strong> is the sum of pass ratios of all the classes divided by the number of the classes.</p>

<p>Return <em>the <strong>maximum</strong> possible average pass ratio after assigning the </em><code>extraStudents</code><em> students. </em>Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.</p>

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

<pre>
<strong>Input:</strong> classes = [[1,2],[3,5],[2,2]], <code>extraStudents</code> = 2
<strong>Output:</strong> 0.78333
<strong>Explanation:</strong> You can assign the two extra students to the first class. The average pass ratio will be equal to (3/4 + 3/5 + 2/2) / 3 = 0.78333.
</pre>

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

<pre>
<strong>Input:</strong> classes = [[2,4],[3,9],[4,5],[2,10]], <code>extraStudents</code> = 4
<strong>Output:</strong> 0.53485
</pre>

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

<ul>
	<li><code>1 &lt;= classes.length &lt;= 10<sup>5</sup></code></li>
	<li><code>classes[i].length == 2</code></li>
	<li><code>1 &lt;= pass<sub>i</sub> &lt;= total<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= extraStudents &lt;= 10<sup>5</sup></code></li>
</ul>

import heapq

class Solution:
    def maxAverageRatio(self, classes: List[List[int]], extraStudents: int) -> float:

        # we could always take the student with the lowest denominator using a heap
        classes = [(num/denom - (num+1)/(denom+1), num, denom) for num, denom in classes]

        # make the heap of classes according to smallest denominator and biggest numerator
        heapq.heapify(classes)

        # check the trivial case that increasing helps nothing
        if classes[0][0] == 0:
            return 1

        # assign the students
        for _ in range(extraStudents):
            
            # get the class with the smallest denominator and the biggest numerator
            _, num, denom = heappop(classes)

            # increase the amount of students and push it back into the heap
            heappush(classes, ((num+1)/(denom+1) - (num+2)/(denom+2), num+1, denom+1))
        return sum(num/denom for _, num, denom in classes)/len(classes)

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