1792. Maximum Average Pass Ratio - Detailed Explanation

Complexity Analysis

• Brute Force Greedy (Approach 1): Let n be the number of classes and m be the number of extra students.

  • Time Complexity: O(n * m). For each of the m extra students, we scan through up to n classes to find the best gain. In the worst case, with both n and m up to 100k, this could be on the order of 10^10 operations, which is not tractable.
  • Space Complexity: O(1) (or O(n) if we count storing the class list and some minor overhead). We only use a few variables for tracking indices and gains. The class data can be considered input space. No significant extra space is used besides that.

• Optimized Greedy with Heap (Approach 2):

  • Time Complexity: O((n + m) log n). We initially push n classes into the heap (cost ~O(n) for building the heap). Then we perform m iterations, each doing a poll and offer on the heap, which are O(log n) each. Thus, roughly O(n + 2m * log n) which simplifies to O((n + m) log n) in big-O terms. With n, m up to 100k, this is on the order of (200k * log(100k)) which is around 200k * ~17 ≈ 3.4 million operations, easily doable.

  • Space Complexity: O(n). We store all classes in the heap. Aside from that, we use a constant amount of extra variables. So it’s linear in the number of classes (which in the worst case is 100k, fine for memory).

Comparison: The optimized approach is significantly better for large inputs. The brute force may be conceptually simpler, but it won’t run within time limits for large constraints. The heap-based approach handles the maximum input sizes efficiently.

Step-by-Step Walkthrough (Optimized Solution)

Let’s do a detailed walkthrough of the optimized greedy approach using a concrete example to see how the algorithm progresses at each step. We’ll use Example 1: classes = [[1,2], [3,5], [2,2]], extraStudents = 2.

Initial State:
We start by calculating the initial pass ratios and the initial gains for each class if one student is added:

We then create a max-heap of these gains. Initially, the heap (max at top) would contain:
[(+0.1667, Class1), (+0.0667, Class2), (+0.0000, Class3)] (We keep track of which class each gain belongs to.)

Iteration 1 (Assign first extra student):

• We extract the top of the heap, which is Class1 with gain 0.1667. This means currently, Class1 is the best place to put an extra student.
• Assign the extra student to Class1. Class1’s stats update from [1,2] to [2,3].
• Now recalculate Class1’s new ratio and new gain:

  • New ratio for Class1 = 2/3 ≈ 0.6667.

  • If we were to add another student to Class1, the ratio would be 3/4 = 0.75. The new gain for an additional student would be 0.75 - 0.6667 = 0.0833.

• We push Class1 back into the heap with its updated gain 0.0833. Classes 2 and 3 remain in the heap with their previous gains (they haven’t changed yet because we didn’t add to them).
• Heap after 1st assignment: [(+0.0833, Class1), (+0.0667, Class2), (+0.0000, Class3)]. Class1 is still on top, but notice its gain dropped from 0.1667 to 0.0833 after getting one student, reflecting diminishing returns.

Iteration 2 (Assign second extra student):

• Again, extract the top of the heap. Now it’s Class1 with gain 0.0833 (it’s still the highest gain; Class2 is 0.0667, Class3 is 0).
• Assign the next extra student to Class1. Update Class1 from [2,3] to [3,4].
• Recalculate Class1’s new ratio and gain:
  • New ratio = 3/4 = 0.75.
  • If one more student were added, ratio would be 4/5 = 0.8. New gain would be 0.8 - 0.75 = 0.0500.
• Push Class1 back into heap with gain 0.0500. (If we had more extra students, we’d continue, but now we’ve used both extras.)
• Heap after 2nd assignment: [(+0.0667, Class2), (+0.0500, Class1), (+0.0000, Class3)]. (Class2 is now at top, but we have no more students to assign.)

Final Calculation:
Now that all extra students are assigned, we compute the average pass ratio:

• Class1: 3/4 = 0.75
• Class2: 3/5 = 0.6 (unchanged from start in this example)
• Class3: 2/2 = 1.0 (unchanged)
  Average = (0.75 + 0.6 + 1.0) / 3 = 2.35 / 3 = 0.78333, which matches the expected result.

Through this walkthrough, we saw:

• How the gain values changed after each assignment (Class1’s gain went down each time it got an extra student).
• How the algorithm always picked the class with the highest current gain.
• How other classes stayed in the heap and would become the top if their gain ever surpassed the others (in this case, if we had more students, eventually Class2’s turn would come once Class1’s gain dropped below Class2’s gain).

This step-by-step process demonstrates the greedy allocation in action, using the priority queue to always make the locally optimal choice (which leads to the globally optimal result).

Common Mistakes & Edge Cases

• Choosing the highest current ratio instead of highest gain: A common pitfall is to mistakenly add students to the class with the lowest pass percentage or the highest fail rate, or conversely the highest current ratio. In truth, neither of those naively yields the optimal result. It’s all about the marginal improvement. For example, a class with a low ratio might not always give the biggest boost if its total is huge (adding one student to a very large class has little effect on its ratio). Always compute the gain from adding a student, and use that to decide.

• Not updating the gain after each assignment: If you add an extra student to a class and then forget to recalculate that class’s new gain for the next iteration, you’ll likely assign subsequent students incorrectly. The gain values change after each student is added to a class (diminishing returns).

• Integer division vs floating point: When implementing, especially in languages like Java or C++, make sure to use floating-point division for calculating ratios and gains. Using integer division (which truncates) will lead to incorrect comparisons of gains (e.g., 1/2 would be 0 in integer math). In Python this isn’t an issue since / yields float by default.

• Precision considerations: The final result needs to be within 1e-5 of the correct answer. Using double precision (double in Java, Python float) is typically sufficient. Be careful if you try to format or round the result – it’s usually fine to just return the double. Summing many ratios (up to 100k classes) in double precision is okay, but avoid using single precision floats as they may introduce too much error.

• Edge Case – No extra students: If extraStudents = 0, simply compute and return the current average pass ratio without any changes.

• Edge Case – All students already passing: If every class already has pass_i = total_i (100% pass rate in each class), then the average is 1.0 and adding extra students won’t change any ratios (they’ll also pass, keeping each class at 100%). The algorithm should handle this by essentially doing nothing (each class’s gain will be 0, so it doesn’t matter where extras go).

• Edge Case – Classes with 1 student or very small classes: The formula works just fine even if a class has only one student. For example, a class [0,1] (0 out of 1 passing) has ratio 0; adding a student yields [1,2] with ratio 0.5 (gain 0.5, which is quite high). The algorithm naturally handles these cases, tending to allocate to small classes if they have low pass counts because the gain is often large.

• Large inputs: Ensure your solution is efficient (use the heap approach). With up to 100k extra students, a slow approach will time out. Also ensure your heap operations and memory usage are optimal. Python’s heapq and Java’s PriorityQueue can handle 100k elements easily.

Alternative Variations

• Maximizing Weighted Average: A slight variation is if we wanted to maximize the total pass ratio or a weighted average where classes have different importance. The greedy strategy would change if, say, classes had different weights. In this problem, each class contributes equally to the average, which is why we simply average the ratios. If instead we cared about the overall pass percentage of all students (combined pass/combined total), the problem would be trivial – you’d always add all extra students to any classes (it wouldn’t matter, as each extra student adds the same amount to total passes overall). It’s the averaging per class that makes it non-trivial.

• Probabilistic variant: If those extra students weren’t guaranteed to pass but instead had some probability p of passing, the problem would turn into maximizing expected value. The greedy approach might still apply by comparing expected gains, but the calculation of “gain” would be different (it might involve diminishing returns in expected pass probability). That would be a more complex variation and not as straightforward as this guaranteed pass scenario.

• Resource Allocation Problems: This problem is essentially a resource allocation task with diminishing returns. Similar setups could be allocating limited resources (like budget, time, or staff) to projects to maximize overall success rate or outcome. The technique we used – always assign the next resource to the project/class that yields the biggest improvement – is a common greedy strategy in such scenarios.

• Greedy + Heap pattern: The approach here is a classic example of using a greedy algorithm with a priority queue to incrementally build an optimal solution. This pattern appears in various problems (some listed below). Whenever you have to repeatedly choose the next best option out of many and the state changes in a way that affects what the "best" next option is, a heap is a great tool to keep track of the best candidates efficiently.

Related Problems for Further Practice

To reinforce the concepts from this problem (greedy choice with priority queue, maximizing or minimizing a certain metric by distribution or selection), you can practice the following problems:

• LeetCode 502. IPO: You have a certain initial capital and a list of projects with their profits and capital requirements. You can do at most k projects. This uses a greedy strategy with a max-heap to always pick the most profitable project available within your current capital. (It’s a similar "pick the best next option" using heaps.)

• LeetCode 1642. Furthest Building You Can Reach: You are given heights of buildings and a limited number of bricks and ladders. You must climb from building to building. The greedy solution uses a min-heap to decide where to use ladders (always use ladders on the largest jumps). This is another resource allocation problem where you greedily allocate a resource to the most costly needs.

• LeetCode 630. Course Schedule III: You have a list of courses with durations and deadlines, and you want to take as many courses as possible. The solution involves sorting by deadlines and using a max-heap to drop the longest course when needed. This showcases a greedy strategy where a heap is used to efficiently decide which course to drop to accommodate a new one.

• LeetCode 1383. Maximum Performance of a Team: You are selecting engineers with speed and efficiency under certain constraints. The approach uses sorting and a min-heap to maintain the team with the largest performance. It’s a bit different, but it relies on a greedy selection and heap to maintain the optimal combination as you iterate.