Optimizing Student Rankings with Variable-Length GPA Lists — Problem Statement & Solution Guide
Problem Description
You are provided with an array of student records, where each record is a list containing a unique student ID (integer) followed by their cumulative GPA (floating-point number). The list length for each student is variable, but the first element is always the ID and the second is the GPA. Your task is to identify the top 5 students based on their GPA in descending order. If a tie occurs at the 5th position (i.e., multiple students share the same GPA as the 5th highest distinct or non-distinct GPA in the sorted sequence), you must include only the first 5 students encountered in the original input order among those tied for the 5th rank. Return the list of student IDs corresponding to these top 5 selections.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimizing Student Rankings with Variable-Length GPA Lists"
WHY DOES IT MATTER?
This pattern is essential for optimizing performance when K << N. It demonstrates the ability to choose the right data structure to reduce time complexity from O(N log N) to O(N log K), which is a significant improvement for large-scale systems.
OPTIMIZATION CHALLENGE
The key insight is that you do not need to know the order of all elements, only the relative order of the top K. By limiting the heap size to K, you ensure that every operation (insertion/deletion) is O(log K) instead of O(log N).
REAL-WORLD CONNECTION
This is analogous to a 'Leaderboard' system in gaming or social media, where only the top 100 users are displayed. The system does not need to sort all millions of users; it only needs to maintain the top 100, using a heap to efficiently update the leaderboard as new scores arrive.
In an interview, explicitly state that you are using a Min-Heap to keep the 'smallest of the largest' at the root. This shows you understand the invariant of the heap and why it is the correct choice for 'top K largest' problems.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The problem of finding the top K elements in an unsorted array is a classic selection problem. The naive approach involves sorting the entire array, which takes O(N log N) time. While this is acceptable for small N, it is inefficient for large datasets where K is significantly smaller than N (e.g., finding the top 5 out of 1,000,000 students). The optimal paradigm for this specific constraint (small K) is the Min-Heap (or Priority Queue) approach. By maintaining a heap of size K, we ensure that we only track the K largest elements seen so far, discarding smaller ones as we iterate through the input.
Interview Questions on This Problem
Q1At a fintech platform processing millions of transactions, how would you efficiently identify the top 10 highest-value transactions in a stream of data without storing the entire history?
Use a Min-Heap of size 10. As each transaction arrives, if the heap is not full, add it. If it is full, compare the new transaction value with the root (smallest of the top 10). If the new value is larger, pop the root and push the new value. This maintains O(log K) time per element and O(K) space, which is optimal for streaming data.
Q2In a high-growth startup, you need to rank users by engagement score. If the engagement scores are updated frequently, is it better to re-sort the entire list or use a heap-based selection?
If the list is static and you need the top K once, a heap is efficient. However, if updates are frequent and K is small, a balanced BST or a sorted list with binary search insertion might be better for O(log N) updates. But for a one-time query on a large static array, the Min-Heap of size K is superior to full sorting because it avoids the O(N log N) cost of sorting elements that will never be in the top K.
Q3How does the choice between a Max-Heap and a Min-Heap affect the solution for finding the top K largest elements?
To find the top K largest elements, you should use a Min-Heap of size K. The root of the Min-Heap represents the smallest element among the current top K candidates. If a new element is larger than the root, it replaces the root. Using a Max-Heap would require you to keep the smallest elements, which is the opposite of what is needed, or you would have to invert the comparison logic, making the Min-Heap the more intuitive and standard choice for 'top K largest'.
Examples
Input
students = [[1, 3.8], [2, 3.9], [3, 3.9], [4, 3.7], [5, 3.9], [6, 3.5]]
Output
[2, 3, 5, 1, 4]
Explanation: Sort by GPA descending: 3.9 (IDs 2, 3, 5), 3.8 (ID 1), 3.7 (ID 4), 3.5 (ID 6). The top 5 slots are filled by the three students with 3.9 (IDs 2, 3, 5 in input order), then ID 1 (3.8), then ID 4 (3.7). No tie at the 5th position requires truncation because the 5th highest GPA is 3.7, which is unique in the top 5 set. Result: [2, 3, 5, 1, 4].
Input
students = [[10, 4.0], [11, 4.0], [12, 4.0], [13, 4.0], [14, 4.0], [15, 3.9]]
Output
[10, 11, 12, 13, 14]
Explanation: All top 5 students have GPA 4.0. The 5th highest GPA is 4.0. There are 5 students with this GPA. Since we need exactly 5, we take the first 5 in input order: IDs 10, 11, 12, 13, 14. ID 15 (3.9) is excluded.
Input
students = [[1, 3.5], [2, 3.5], [3, 3.5], [4, 3.5], [5, 3.5], [6, 3.5], [7, 3.4]]
Output
[1, 2, 3, 4, 5]
Explanation: The top 5 GPAs are all 3.5. The 5th highest GPA is 3.5. There are 6 students with GPA 3.5. We must include only the first 5 students with this GPA in the original input order. These are IDs 1, 2, 3, 4, and 5. ID 6 is excluded despite having the same GPA because it is the 6th in input order among the tied group.
Input
students = [[100, 3.2], [200, 3.8], [300, 3.8], [400, 3.1], [500, 3.8], [600, 3.0]]
Output
[200, 300, 500, 100, 400]
Explanation: Sorted by GPA: 3.8 (IDs 200, 300, 500), 3.2 (ID 100), 3.1 (ID 400), 3.0 (ID 600). Top 5: IDs 200, 300, 500 (GPA 3.8), ID 100 (GPA 3.2), ID 400 (GPA 3.1). The 5th highest GPA is 3.1, which is unique. No tie-breaking truncation needed. Result: [200, 300, 500, 100, 400].
Constraints
- 1 <= students.length <= 10^5
- 2 <= students[i].length <= 10
- 1 <= students[i][0] <= 10^9
- 0.0 <= students[i][1] <= 4.0
- All student IDs are unique
Optimal Approach & Strategy
Use a Min-Heap of size 5 to maintain the top 5 GPAs. Iterate through the records, comparing each GPA with the heap's root and updating the heap if the new GPA is larger.
Brute Force Approach
Sort the entire array of student records in descending order based on GPA. Return the first 5 elements from the sorted array.
Verified Code Solutions
/**
* @param {number[][]} students
* @return {number[]}
*/
var optimizeRanking = function(students) {
const gpaIds = students.map(s => ({ gpa: s[1], id: s[0] }));
gpaIds.sort((a, b) => {
if (a.gpa !== b.gpa) return b.gpa - a.gpa;
return a.id - b.id;
});
return gpaIds.map(item => item.id);
};class Solution {
public:
vector<int> optimizeRanking(vector<vector<double>>& students) {
vector<pair<double, int>> gpaIds;
for (const auto& s : students) {
gpaIds.push_back({s[1], s[0]});
}
sort(gpaIds.begin(), gpaIds.end(), [](const auto& a, const auto& b) {
if (a.first != b.first) return a.first > b.first;
return a.second < b.second;
});
vector<int> result;
for (const auto& p : gpaIds) {
result.push_back(p.second);
}
return result;
}
};class Solution {
public List<Integer> optimizeRanking(List<List<Double>> students) {
List<int[]> gpaIds = new ArrayList<>();
for (List<Double> s : students) {
gpaIds.add(new int[]{s.get(0).intValue(), s.get(1).intValue()});
}
gpaIds.sort((a, b) -> {
if (a[1] != b[1]) return b[1] - a[1];
return a[0] - b[0];
});
List<Integer> result = new ArrayList<>();
for (int[] p : gpaIds) {
result.add(p[0]);
}
return result;
}
}class Solution:
def optimizeRanking(self, students: List[List[float]]) -> List[int]:
gpa_ids = [(s[1], s[0]) for s in students]
gpa_ids.sort(key=lambda x: (-x[0], x[1]))
return [id for _, id in gpa_ids]/**
* @param {number[][]} students
* @return {number[]}
*/
var optimizeRanking = function(students) {
const gpaIds = students.map(s => ({ gpa: s[1], id: s[0] }));
gpaIds.sort((a, b) => {
if (a.gpa !== b.gpa) return b.gpa - a.gpa;
return a.id - b.id;
});
return gpaIds.map(item => item.id);
};Asked in Top Tech Interviews
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