Vault Registry Evaluator 43 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and registry metrics and a threshold K, construct an optimal algorithm to evaluate and compute the target evaluator value by summing all numbers greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Registry Evaluator 43"
WHY DOES IT MATTER?
Filtering and aggregating large data streams in one pass is a cornerstone of high‑throughput systems.
OPTIMIZATION CHALLENGE
Eliminating sorting or nested loops drops the complexity from O(n log n) or O(n²) to linear time.
REAL-WORLD CONNECTION
Think of monitoring server metrics where you need the total load of all nodes exceeding a safety threshold.
Always profile for early exit opportunities—if K is larger than the maximum element, you can skip the entire scan.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a linear scan over an array of vault and registry metrics, applying a simple predicate (value > K) and aggregating qualifying elements. This fits the classic "filter‑and‑reduce" paradigm, where the filter step discards irrelevant data and the reduce step combines the remaining values into a single sum. Naïve alternatives—such as sorting the entire list or using nested loops to compare each element against every other—inflate time complexity to O(n log n) or O(n²), which is prohibitive for the massive streams typical in telemetry systems. The optimal approach leverages the fact that the predicate is independent of element order, allowing a single pass with constant auxiliary storage, achieving O(n) time and O(1) extra space. This aligns with the broader optimal paradigm of streaming algorithms: process each item once, maintain minimal state, and avoid unnecessary data structures.
Interview Questions on This Problem
Q1How would you handle the case where the input sequence is provided as a lazy iterator rather than a concrete array?
Treat the iterator as a stream and apply the same filter‑and‑sum logic while iterating. Since you never store the whole sequence, memory usage stays O(1).
Q2What modifications are needed if the threshold K can change dynamically during the scan?
You would need to maintain a data structure (e.g., a balanced BST) to support re‑evaluation of previously seen elements, raising complexity to O(log n) per update. In most real‑time systems, recomputing the sum after each K change is avoided by batching or resetting the scan.
Q3Can you extend the solution to also count how many elements exceed K while summing them?
Yes, maintain a second counter variable that increments whenever a value > K is encountered. Both sum and count are updated in the same O(n) pass.
Examples
Input
[10, 20, 30, 40, 50], K = 30
Output
120
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K = 30, we first filter out numbers less than or equal to K, giving [40, 50]. Then, we add these numbers to get the target evaluator value, which is 40 + 50 = 90.
Input
[100, 200, 300, 400, 500], K = 250
Output
1200
Explanation: Step-by-step: with input [100, 200, 300, 400, 500] and K = 250, we first filter out numbers less than or equal to K, giving [300, 400, 500]. Then, we add these numbers to get the target evaluator value, which is 300 + 400 + 500 = 1200.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal solution scans once, checks the predicate, and updates a running sum, achieving O(n) time and O(1) space.
Brute Force Approach
A brute force method would sort the array then sum the tail, costing O(n log n) time and extra space for the sorted copy.
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): sum = 0; for num in nums: if num > K: sum += num; return sumfunction 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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