1792. Maximum Average Pass Ratio

Problem

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 classes, where classes[i] = [passᵢ, totalᵢ]. You know beforehand that in the iᵗʰ class, there are totalᵢ total students, but only passᵢ number of students will pass the exam.

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

The pass ratio 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 average pass ratio is the sum of pass ratios of all the classes divided by the number of the classes.

Return the maximum possible average pass ratio after assigning the extraStudents students. Answers within 10⁻⁵ of the actual answer will be accepted.

https://leetcode.com/problems/maximum-average-pass-ratio/

Example 1:

Input: classes = [[1,2],[3,5],[2,2]], extraStudents = 2
Output: 0.78333
Explanation: 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.

Example 2:

Input: classes = [[2,4],[3,9],[4,5],[2,10]], extraStudents = 4
Output: 0.53485

Constraints:

1 <= classes.length <= 10⁵
classes[i].length == 2
1 <= passᵢ <= totalᵢ <= 10⁵
1 <= extraStudents <= 10⁵

Test Cases

1 2	class Solution: def maxAverageRatio(self, classes: List[List[int]], extraStudents: int) -> float:

solution_test.py

import pytest

from solution import Solution


@pytest.mark.parametrize('classes, extraStudents, expected', [
    ([[1,2],[3,5],[2,2]], 2, 0.78333),
    ([[2,4],[3,9],[4,5],[2,10]], 4, 0.53485),
])
class Test:
    def test_solution(self, classes, extraStudents, expected):
        sol = Solution()
        assert sol.maxAverageRatio(classes, extraStudents) == pytest.approx(expected, abs=1e-5)

Thoughts

求最大的平均通过率，也可以先计算最小的平均不通过率。记 pᵢ = passᵢ、tᵢ = totalᵢ、fᵢ = totalᵢ - passᵢ，记给第 i 个班加入 eᵢ 个「必过学生」，那么：

\begin{array}{rl} & \max_{\{e_1,e_2,\dots,e_n\}}\bar{pr} \\ = & \frac{1}{n}\max_{\{e_1,e_2,\dots,e_n\}}\sum_{i=1}^{n}{(p_i+e_i)/(t_i+e_i)} \\ = & 1-\frac{1}{n}\min_{\{e_1,e_2,\dots,e_n\}}\sum_{i=1}^{n}{f_i/(t_i+e_i)} \end{array}

对于任何一个班，只要 fᵢ ≠ 0，加入「必过学生」就一定能提高通过率（降低不通过率）。关键是给每个班分配多大的 eᵢ。

想看看有没有更「数学」的方法，但没推导出来。只好对「必过学生」一个一个地选择放到哪个班。

假设把一个「必过学生」放进某个班，看能降低多少不通过率，即 $f_i/t_i-f_i/(t_i+1)$ 。对所有的班级，取这个值最大的即可。

用最大堆记录所有班「加入一个必过学生能降低多少不通过率」，每次取堆顶的班，加入一个「必过学生」后再重新放入堆中。

时间复杂度 O((n + k) log n)，其中 k = extraStudents。空间复杂度 O(n)。

Code

solution.py

from heapq import heapify, heapreplace


class Solution:
    def maxAverageRatio(self, classes: list[list[int]], extraStudents: int) -> float:
        n = len(classes)
        queue = [((t-p)/(t+1) - (t-p)/t, t - p, t) for p, t in classes]
        heapify(queue)
        for _ in range(extraStudents):
            _, f, t = queue[0]
            t += 1
            heapreplace(queue, (f/(t+1) - f/t, f, t))

        return 1 - sum(f/t for _, f, t in queue) / n

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。