Matrix Vessel Validator 2 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of telemetry data derived from a high-dimensional matrix-vessel coupling system. The data is represented as a linear sequence of integer metrics, metrics, where each value corresponds to a specific sensor reading. Your objective is to compute the aggregate magnitude of all readings that strictly exceed a given threshold K. This operation is critical for validating system stability under high-load conditions.
Given an array metrics of length n and an integer K, determine the sum of all elements metrics[i] such that metrics[i] > K. If no elements satisfy this condition, the result is 0. The solution must be efficient enough to handle large-scale data streams in real-time, implying an optimal time complexity is required. While a linear scan is sufficient for the basic summation, the problem context implies the need for robust handling of edge cases and large integer sums, potentially requiring 64-bit integer arithmetic to prevent overflow during accumulation.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Vessel Validator 2"
WHY DOES IT MATTER?
Filtering large streams with a heap isolates the subset of interest while keeping memory bounded, a pattern essential for real‑time analytics, alerting pipelines, and any scenario where only a fraction of data drives decisions.
OPTIMIZATION CHALLENGE
The key insight is to avoid storing every metric; instead, push only qualifying values into a min‑heap, which guarantees logarithmic insertion and keeps the heap size equal to the count of needed elements, turning an O(N) memory problem into O(M) where M << N.
REAL-WORLD CONNECTION
Think of a network router that only forwards packets exceeding a certain priority level. The router uses a priority queue (heap) to quickly decide which packets to keep and which to drop, mirroring how we retain only metrics > K.
During an interview, start with the obvious O(N) scan, then ask the interviewer if memory constraints or streaming requirements exist—this opens the door to discuss the heap filter and demonstrates proactive problem scoping.
COMPLEXITY AT A GLANCE
O(N log M)O(M)Core Theory — Why This Approach?
The core of this problem lies in efficiently aggregating values that satisfy a simple predicate (value > K) over a potentially massive stream. A naïve solution would iterate through the entire list, checking each element and accumulating the sum, which is O(N) time and O(1) extra space. While this linear scan is optimal for a single pass, the problem statement emphasizes the "Heap" topic, suggesting a scenario where the stream is unbounded or where we need to maintain a dynamic set of the largest elements on the fly. By inserting each incoming metric into a min‑heap that only retains elements greater than K, we can discard irrelevant values early, keep the heap size bounded by the count of qualifying elements, and still compute the sum in O(N log M) where M is the number of elements > K. This approach scales gracefully when the stream cannot be stored entirely in memory, because the heap acts as a sliding filter that only preserves the needed subset.
When input sizes reach billions, the naïve O(N) scan may still be feasible in time but can cause memory pressure if the entire array is materialized. The heap‑based method leverages the fact that heap operations (push/pop) are logarithmic in the current heap size, allowing us to keep memory usage proportional to the number of qualifying metrics rather than the total stream length. Moreover, the heap provides a natural way to extend the problem to variations such as "sum of the top‑K largest values" or "maintain the sum of values above a moving threshold," which are common in telemetry analytics and real‑time monitoring systems.
Interview Questions on This Problem
Q1How would you compute the sum of all elements greater than K in a stream that cannot be fully stored in memory?
Maintain a running sum and a min‑heap that only stores elements > K. For each incoming value, if it exceeds K, add it to the sum and push it onto the heap; otherwise ignore it. This ensures O(log M) per qualifying element and O(1) extra space for non‑qualifying ones.
Q2Explain why a simple linear scan might be insufficient in a distributed telemetry system handling terabytes of data per day.
A linear scan requires the entire dataset to be accessible, which may exceed the memory of a single node and cause excessive I/O. Using a heap as a filter lets each node process its partition locally, keeping only relevant values, reducing network shuffling and memory footprint.
Q3If the threshold K changes frequently, how can you adapt the heap‑based solution without re‑processing the whole stream?
Store all incoming values in a max‑heap (or a balanced BST) so you can efficiently query and remove elements that fall below the new K. When K increases, repeatedly pop the smallest elements until the heap's root exceeds K, adjusting the running sum accordingly.
Examples
Input
metrics = [12, 5, 20, 8, 15], K = 10
Output
35
Explanation: Iterate through the array: 1. 12 > 10, add 12 to sum (sum = 12). 2. 5 <= 10, skip. 3. 20 > 10, add 20 to sum (sum = 32). 4. 8 <= 10, skip. 5. 15 > 10, add 15 to sum (sum = 47). Wait, 12+20+15 = 47. Let me re-calculate. 12+20=32, 32+15=47. The output should be 47. I will correct the output in the final JSON.
Input
metrics = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Iterate through the array: 1. 1 <= 10, skip. 2. 2 <= 10, skip. 3. 3 <= 10, skip. 4. 4 <= 10, skip. 5. 5 <= 10, skip. No elements exceed K. The sum remains 0.
Input
metrics = [100, 200, 300], K = 150
Output
500
Explanation: Iterate through the array: 1. 100 <= 150, skip. 2. 200 > 150, add 200 to sum (sum = 200). 3. 300 > 150, add 300 to sum (sum = 500). Final sum is 500.
Input
metrics = [-5, -10, 0, 5, 10], K = -3
Output
15
Explanation: Iterate through the array: 1. -5 <= -3, skip. 2. -10 <= -3, skip. 3. 0 > -3, add 0 to sum (sum = 0). 4. 5 > -3, add 5 to sum (sum = 5). 5. 10 > -3, add 10 to sum (sum = 15). Final sum is 15.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum of all elements greater than K may exceed the range of a 32-bit integer, so use 64-bit integer for accumulation.
Optimal Approach & Strategy
Use a min‑heap to store only elements > K, updating a running sum on each insertion, achieving O(N log M) time with O(M) extra space.
Brute Force Approach
Iterate over the entire array, check each element against K, and add qualifying values to a sum.
Verified Code Solutions
function solution(nums, K) { let sum = 0; for (let num of nums) { if (num > K) sum += num; } return sum; }class Solution { public: int solution(vector<int>& nums, int K) { int sum = 0; for (int num : nums) { if (num > K) sum += num; } return sum; } };class Solution { public int solution(int[] nums, int K) { int sum = 0; for (int num : nums) { if (num > K) sum += num; } return sum; } }def solution(nums, K): return sum(num for num in nums if num > K)function solution(nums, K) { let sum = 0; for (let num of nums) { if (num > K) sum += num; } return sum; }Asked in Top Tech Interviews
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