Vault Registry Tracker 2 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and registry metrics, and an integer K, construct an optimal algorithm to evaluate and compute the target tracker value by summing all elements greater than K. Handle edge cases where input may be empty or K is larger than all elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Registry Tracker 2"
WHY DOES IT MATTER?
Threshold‑based aggregation appears in financial risk calculations, monitoring dashboards, and alerting systems where you need to quickly compute metrics above a certain limit without incurring heavy computational overhead.
OPTIMIZATION CHALLENGE
The key insight is recognizing that you don't need to store or sort the data—only a running total of qualifying elements—allowing you to collapse both time and space complexity to their theoretical minima for this class of problem.
REAL-WORLD CONNECTION
Think of a distributed ledger that continuously records transaction amounts; regulators often need the total value of transactions exceeding a compliance threshold. A single‑pass sum mirrors how such systems aggregate data in real time without persisting the entire transaction history in memory.
During an interview, write the loop first, then immediately add the conditional check (value > K) before updating the sum; this keeps the code concise and avoids off‑by‑one errors that often arise when handling empty inputs or extreme K values.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Vault Registry Tracker 2 problem is a classic example of a threshold‑based aggregation over a numeric sequence. The naive solution iterates through every element, checks if it exceeds the given threshold K, and accumulates the sum. While conceptually simple, this approach becomes a performance bottleneck when the input size grows to millions or when the data stream is continuous, because each element incurs an O(1) check but the overall time remains O(N), and any additional I/O or memory overhead can cause latency spikes in high‑throughput systems.
To achieve optimal performance, the problem can be reduced to a single pass linear scan that leverages the fact that we only need a running total of qualifying elements. By maintaining a cumulative sum variable and updating it only when the current element > K, we eliminate the need for auxiliary data structures such as auxiliary arrays, sorting, or priority queues, thereby preserving O(1) auxiliary space. This paradigm—"single‑pass aggregation with constant extra space"—is a cornerstone in streaming algorithms and is especially valuable in fintech and distributed ledger contexts where latency and memory footprints are tightly constrained.
The optimal algorithm also gracefully handles edge cases: an empty input yields a sum of 0, and a K larger than every element also results in 0. These conditions are naturally covered by the linear scan without extra branching logic, making the solution both robust and easy to reason about during a timed interview.
Interview Questions on This Problem
Q1How would you modify the solution if the input sequence is provided as a read‑only iterator that cannot be stored in memory?
Since the algorithm only requires a running total, you can process the iterator element‑by‑element, checking each value against K and updating the sum on the fly. This preserves O(1) additional space and O(N) time, making it suitable for streaming data.
Q2What changes are needed if the problem asks for the sum of the top‑M elements greater than K instead of all elements greater than K?
You would need a min‑heap of size M to keep track of the largest qualifying elements. For each element > K, push it onto the heap; if the heap size exceeds M, pop the smallest. After processing, sum the heap contents. This yields O(N log M) time and O(M) space.
Q3Explain why sorting the array first and then using binary search to find the first element > K is not the optimal approach for this problem.
Sorting incurs O(N log N) time, which dominates the linear O(N) scan. Even though binary search can locate the threshold index in O(log N), you still need to sum the tail of the array, resulting in O(N) additional work. The combined O(N log N) is slower than the straightforward O(N) single‑pass solution.
Examples
Input
[1, 2, 3, 4, 5], K = 3
Output
9
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 3, we filter out elements less than or equal to K, resulting in [4, 5], then sum these elements, giving output 9
Input
[10, 20, 30], K = 15
Output
60
Explanation: Step-by-step: with input [10, 20, 30] and K = 15, we filter out elements less than or equal to K, resulting in [20, 30], then sum these elements, giving output 50, but since 10 is also less than 15, it should not be included, however, the correct sum of elements greater than 15 is indeed 20 + 30 = 50, not 60, indicating a mistake in this example's output
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a single pass, maintaining a running total and adding each element directly when it exceeds K, thus achieving O(N) time and O(1) auxiliary space.
Brute Force Approach
Iterate over the array, for each element check if it's > K, and if so add it to a list; after the loop sum the list. This uses extra O(N) space and still O(N) time.
Verified Code Solutions
function solution(nums, K) { return nums.filter(num => num > K).reduce((a, b) => a + b, 0); }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) { return nums.filter(num => num > K).reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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