1792. Maximum Average Pass Ratio

Method 1

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.

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

Method 1

【O((n+k)log(n) time | O(n) space】

package Leetcode.Greedy;

import java.util.Comparator;
import java.util.PriorityQueue;

/**
 * @author zhengxingxing
 * @date 2024/12/15
 */
public class MaximumAveragePassRatio {
    public static double maxAverageRatio(int[][] classes, int extraStudents) {
        int len = classes.length;

        // Custom comparator to prioritize classes based on "marginal gain" in pass ratio
        PriorityQueue<double[]> pq = new PriorityQueue<double[]>(
                new Comparator<double[]>(){
                    @Override
                    public int compare(double[] a, double[] b ){
                        // Calculate the improvement in pass ratio when adding one student
                        // For class `a`: (a[0]+1)/(a[1]+1) - a[0]/a[1]
                        // Example: If a class improves from 1/2 to 2/3:
                        // Original ratio: 1/2 = 0.5
                        // New ratio: 2/3 ≈ 0.667
                        // Gain: 0.667 - 0.5 = 0.167
                        double diffa = (a[0]+1)/(a[1]+1) - a[0]/a[1];
                        double diffb = (b[0]+1)/(b[1]+1) - b[0]/b[1];

                        // Comparison logic:
                        // 1. If the gain is equal, return 0
                        if (diffa == diffb) {
                            return 0;
                        }
                        // 2. If the gain for class `a` is greater than `b`, return -1
                        // PriorityQueue is a min-heap, so returning -1 means `a` has higher priority
                        // We want the class with the maximum gain to be at the front
                        return diffa > diffb ? -1 : 1;
                    }
                }
        );
        // Add all class information to the priority queue
        for (int i = 0; i < len; i++) {
            pq.offer(new double[] {
                    Double.valueOf(classes[i][0]),
                    Double.valueOf(classes[i][1])
            });
        }

        // Allocate additional students
        while (!pq.isEmpty() && extraStudents > 0) {
            double[] p = pq.poll(); // Remove the class with the highest marginal gain

            p[0] += 1; // Increment passed students
            p[1] += 1; // Increment total students

            pq.offer(p); // Re-add the updated class back to the queue
            extraStudents--;
        }

        // Calculate the final average pass ratio
        double res = 0;
        for (double[] p : pq) {
            res += p[0] / p[1];
        }

        // Return the average pass ratio
        return res / Double.valueOf(len);
    }

    public static void main(String[] args) {

        // Test case 1: Three classes, two extra students
        int[][] classes1 = { { 1,2},{3,5},{2,2 } };
        int extraStudents1 = 2;
        double result1 = maxAverageRatio(classes1, extraStudents1);
        System.out.println("Test case 1 result: " + result1);

        // Test case 2: Four classes, four extra students
        int[][] classes2 = { { 2,4},{3,9},{4,5},{2,10 } };
        int extraStudents2 = 4;
        double result2 = maxAverageRatio(classes2, extraStudents2);
        System.out.println("Test case 2 result: " + result2);

        // Boundary test case: Only one class
        int[][] classes3 = { { 1,3 } };
        int extraStudents3 = 1;
        double result3 = maxAverageRatio(classes3, extraStudents3);
        System.out.println("Test case 3 result: " + result3);
    }
}

Method 1

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