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1792. Maximum Average Pass Ratio

Description

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

 

Constraints:

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

Solutions

Solution 1: Priority Queue (Max-Heap of Increment)

Suppose a class currently has a pass rate of $\frac{a}{b}$. If we arrange a smart student into this class, then the pass rate of the class will become $\frac{a+1}{b+1}$. We can find that the increment of the pass rate is $\frac{a+1}{b+1} - \frac{a}{b}$.

We maintain a max-heap, which stores the increment of the pass rate for each class.

Perform extraStudents operations, each time taking a class from the top of the heap, adding $1$ to both the number of students and the number of passes in this class, then recalculating the increment of the pass rate of this class and putting it back into the heap. Repeat this process until all students are allocated.

Finally, we sum up the pass rates of all classes, and then divide by the number of classes to get the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the number of classes.

  • class Solution {
    public double maxAverageRatio(int[][] classes, int extraStudents) {
        PriorityQueue<double[]> pq = new PriorityQueue<>((a, b) -> {
            double x = (a[0] + 1) / (a[1] + 1) - a[0] / a[1];
            double y = (b[0] + 1) / (b[1] + 1) - b[0] / b[1];
            return Double.compare(y, x);
        });
        for (var e : classes) {
            pq.offer(new double[] {e[0], e[1]});
        }
        while (extraStudents-- > 0) {
            var e = pq.poll();
            double a = e[0] + 1, b = e[1] + 1;
            pq.offer(new double[] {a, b});
        }
        double ans = 0;
        while (!pq.isEmpty()) {
            var e = pq.poll();
            ans += e[0] / e[1];
        }
        return ans / classes.length;
    }
}

  • class Solution {
public:
    double maxAverageRatio(vector<vector<int>>& classes, int extraStudents) {
        priority_queue<tuple<double, int, int>> pq;
        for (auto& e : classes) {
            int a = e[0], b = e[1];
            double x = (double) (a + 1) / (b + 1) - (double) a / b;
            pq.push({x, a, b});
        }
        while (extraStudents--) {
            auto [_, a, b] = pq.top();
            pq.pop();
            a++;
            b++;
            double x = (double) (a + 1) / (b + 1) - (double) a / b;
            pq.push({x, a, b});
        }
        double ans = 0;
        while (pq.size()) {
            auto [_, a, b] = pq.top();
            pq.pop();
            ans += (double) a / b;
        }
        return ans / classes.size();
    }
};

  • class Solution:
    def maxAverageRatio(self, classes: List[List[int]], extraStudents: int) -> float:
        h = [(a / b - (a + 1) / (b + 1), a, b) for a, b in classes]
        heapify(h)
        for _ in range(extraStudents):
            _, a, b = heappop(h)
            a, b = a + 1, b + 1
            heappush(h, (a / b - (a + 1) / (b + 1), a, b))
        return sum(v[1] / v[2] for v in h) / len(classes)


  • func maxAverageRatio(classes [][]int, extraStudents int) float64 {
	pq := hp{}
	for _, e := range classes {
		a, b := e[0], e[1]
		x := float64(a+1)/float64(b+1) - float64(a)/float64(b)
		heap.Push(&pq, tuple{x, a, b})
	}
	for i := 0; i < extraStudents; i++ {
		e := heap.Pop(&pq).(tuple)
		a, b := e.a+1, e.b+1
		x := float64(a+1)/float64(b+1) - float64(a)/float64(b)
		heap.Push(&pq, tuple{x, a, b})
	}
	var ans float64
	for len(pq) > 0 {
		e := heap.Pop(&pq).(tuple)
		ans += float64(e.a) / float64(e.b)
	}
	return ans / float64(len(classes))
}

type tuple struct {
	x float64
	a int
	b int
}

type hp []tuple

func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool {
	a, b := h[i], h[j]
	return a.x > b.x
}
func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any)   { *h = append(*h, v.(tuple)) }
func (h *hp) Pop() any     { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }

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