<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>

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

        impacts = [0] * n

        #  step 1, calculate and store tuple data (-impact, passCount, passCount)

        for i in range(n):
            # ic(classes[i])
            passCount = classes[i][0]
            totalCount = classes[i][1]

            # calculate the impact for class i
            currentRatio = passCount/totalCount
            expectedRatioAfterUpdate = (passCount+1)/(totalCount+1)
            impact = expectedRatioAfterUpdate - currentRatio
            # ic(passCount, totalCount, currentRatio, expectedRatioAfterUpdate)

            # now that we have the impact for class i, save it
            impacts[i] = (-impact, passCount, totalCount)
        # ic(impacts)

        #  ------------------------------------

        # Step 2 - Now that we have initial impacts loaded. Lets heapify it

        heapq.heapify(impacts)
        # ic(impacts)

        # -------------------------------------

        #  Now we have easy access to the most impactful class to place a student.
        #  for all the students, we can now walk through and calculate new impacts as we go.

        while extraStudents > 0:

            #  1 - pick greatest impact class
            xImp, xPass, xTotal = heapq.heappop(impacts)

            # ic(extraStudents, xImp, xPass, xTotal)

            # 2 - assign student to that class
            xPass += 1
            xTotal += 1
            #  calculate new ratio
            xRatio = xPass/xTotal
            xNewRatio = (xPass+1)/(xTotal+1)
            impact = xNewRatio - xRatio
            # insert back into the heap

            heapq.heappush(impacts, (-impact, xPass, xTotal))
            # ic(impacts)
            extraStudents -= 1

        # ------------------------------------
        # By here, we have the final version of our impacts table with
        # all the extra students added.
        # We can finally calculate our "maximum possible average pass ratio"
        summedPassRatios = 0
        for yImpact, ypassCount, yTotal in impacts:
            # ic(yImpact, ypassCount, yTotal)
            summedPassRatios += ypassCount/yTotal

        return summedPassRatios/n
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