# Program to find maximum average pass ratio in Python

PythonServer Side ProgrammingProgramming

Suppose we have a list of classes where classes[i] represents [pass_i, total_i] represents the number of students passed the examination of ith class and total number of students of the ith class respectively. We also have another value extra. This indicates extra number of brilliant students that are guaranteed to pass the exam of any class they are assigned to. We have to assign each of the extra students to a class in a way that maximizes the average number of passed student across all the classes. The pass ratio of a class determined by the number of students of the class that will passed divided by the total number of students of the class. And the average pass ratio is the sum of pass ratios of all the classes divided by the number of the classes. We have to find the maximum possible average pass ratio after assigning the extra students.

So, if the input is like classes = [[2,3],[4,6],[3,3]], extra = 3, then the output will be 0.83809, because two extra students at first class and add one extra student to second class to maximize the ratio, so now average is (4/5 + 5/7 + 3/3)/3 = 0.83809.

To solve this, we will follow these steps −

• h := a list of tuples like (a/b-(a + 1)/(b + 1), a, b) for each pair (a, b) in classes

• heapify h

• while extra is non-zero, do

• (v, a, b) := top of h, and delete it from h

• (a, b) := (a + 1, b + 1)

• insert (-(a + 1) /(b + 1) + a / b, a, b) into heap

• extra := extra - 1

• return average from all tuples of h

## Example

Let us see the following implementation to get better understanding −

import heapq
def solve(classes, extra):
h = [(a / b - (a + 1) / (b + 1), a, b) for a, b in classes]
heapq.heapify(h)
while extra:
v, a, b = heapq.heappop(h)
a, b = a + 1, b + 1
heapq.heappush(h, (-(a + 1) / (b + 1) + a / b, a, b))
extra -= 1
return sum(a / b for v, a, b in h) / len(h)

classes = [[2,3],[4,6],[3,3]]
extra = 3
print(solve(classes, extra))

## Input

[[2,3],[4,6],[3,3]], 3

## Output

0