Maximum Average Pass Ratio

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.

Input & Output

Example 1 — Basic Assignment

$ Input: classes = [[1,2],[3,5]], extraStudents = 2

› Output: 0.675

💡 Note: Initially: Class 1 has ratio 1/2=0.5, Class 2 has ratio 3/5=0.6. The improvement from adding one student to Class 1 is (2/3-1/2)=1/6≈0.167, while for Class 2 it's (4/6-3/5)=1/15≈0.067. So we add the first student to Class 1: [2,3]. The new improvements are (3/4-2/3)=1/12≈0.083 for Class 1 and 1/15≈0.067 for Class 2. Class 1 still has better improvement, so we add the second student there too: [3,4]. Final ratios: 3/4=0.75 and 3/5=0.6. Average = (0.75+0.6)/2 = 0.675.

Example 2 — Single Class

$ Input: classes = [[2,4]], extraStudents = 1

› Output: 0.6

💡 Note: Initially: Class has ratio 2/4 = 0.5. Adding one student gives (3,5) with ratio 3/5 = 0.6. Since there's only one class, the average is 0.6.

Example 3 — No Improvement Needed

$ Input: classes = [[3,3],[4,4]], extraStudents = 2

› Output: 1.0

💡 Note: Both classes already have 100% pass rate. Adding students maintains 100% pass rate: (4,4) and (5,5) both have ratio 1.0. Average remains 1.0.

Constraints

1 ≤ classes.length ≤ 10⁵
classes[i].length = 2
1 ≤ pass_i ≤ total_i ≤ 10⁵
1 ≤ extraStudents ≤ 10⁵

Visualization

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Asked in

G Google 28 f Facebook 15 a Amazon 12 M Microsoft 8

The key insight is that we should always assign each extra student to the class that gives the maximum improvement in pass ratio. Using a priority queue to track improvements leads to the optimal greedy solution. Best approach is Greedy with Priority Queue: Time: O(extraStudents × log n), Space: O(n)

Common Approaches

✓ Frequency Count

⏱️ Time: N/A Space: N/A

Brute Force - Try All Distributions

⏱️ Time: O(n^extraStudents) Space: O(1)

Generate all possible ways to distribute extraStudents among the classes and calculate the average pass ratio for each distribution. This requires trying every combination which becomes exponentially expensive.

Greedy with Priority Queue

⏱️ Time: O(extraStudents × log n) Space: O(n)

Use a max-heap to track which class would benefit most from adding one student. The improvement is calculated as the difference in pass ratio before and after adding a student. Always pick the class with maximum improvement.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int min(int a, int b) {
    return a < b ? a : b;
}

int solution(char* text) {
    // Count frequency using array (assuming lowercase letters)
    int freq[26] = {0};
    
    for (int i = 0; text[i] != '\0'; i++) {
        if (text[i] >= 'a' && text[i] <= 'z') {
            freq[text[i] - 'a']++;
        }
    }
    
    // Get counts for required characters
    // 'balloon' needs: 1 b, 1 a, 2 l, 2 o, 1 n
    int bCount = freq['b' - 'a'];
    int aCount = freq['a' - 'a'];
    int lCount = freq['l' - 'a'];
    int oCount = freq['o' - 'a'];
    int nCount = freq['n' - 'a'];
    
    // Calculate maximum balloons
    int result = bCount;
    result = min(result, aCount);
    result = min(result, lCount / 2);
    result = min(result, oCount / 2);
    result = min(result, nCount);
    
    return result;
}

int main() {
    char text[10001];
    fgets(text, sizeof(text), stdin);
    text[strcspn(text, "\n")] = 0;
    
    int result = solution(text);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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