BackhardHeapGoogleAmazon

Vault Registry Detector 40 Solution

Problem Statement

Given a sequence of data elements representing vault and registry metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints. The target detector value is the sum of all elements in the sequence that are greater than K.

Example 1
Input
[6, 7, 8, 9, 10], 5
Output
10

Explanation: Step-by-step: Given the input array [6, 7, 8, 9, 10] and K = 5, we iterate through the array and sum up all elements greater than 5. The elements 6, 7, 8, 9, and 10 are greater than 5, so we add them up to get the final output of 10.

Example 2
Input
[1, 2, 3, 4, 5], 5
Output
0

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and K = 5, we iterate through the array and sum up all elements greater than 5. However, there are no elements greater than 5 in the array, so we return 0 as the final output.

Constraints

  • 1 <= N <= 10^5
  • -10^4 <= metrics[i] <= 10^4
  • 1 <= K <= N
Live Compiler1 Free Run Available
Loading Editor...
Test Cases & Output
Click "Run" to test your 1 free compile trial!

🚀 Practice this problem

Run code, get AI hints & track streak

Sign Up Free

Vault Registry Detector 40 — Problem Statement & Solution Guide

HeapHard2D Grid DP
TimeO(n)
|
SpaceO(1)

Problem Description

Given a sequence of data elements representing vault and registry metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints. The target detector value is the sum of all elements in the sequence that are greater than K.

DSA Pattern Breakdown

DSA Pattern Breakdown

"Vault Registry Detector 40"

hard

WHY DOES IT MATTER?

Linear‑time aggregation is a foundational pattern for processing massive streams efficiently.

OPTIMIZATION CHALLENGE

Eliminating sorting or nested loops reduces the complexity from O(n log n) or O(n^2) to O(n).

REAL-WORLD CONNECTION

Think of a monitoring system that continuously sums metrics exceeding a threshold, like total high‑latency requests.

Always scan once, keep a running total, and avoid extra containers unless you need ordering or random access.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)
💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to a single linear scan where each element is compared against the threshold K and, if larger, added to an accumulator. This is a classic example of a prefix-sum style aggregation that can be solved in O(n) time using constant extra space.

A naive approach might attempt sorting or using nested loops to filter elements, which inflates time complexity to O(n log n) or O(n^2) and is unnecessary because the ordering of elements does not affect the sum. The optimal paradigm leverages the fact that the sum operation is associative and commutative, allowing a single-pass reduction without auxiliary data structures.

Interview Questions on This Problem

Q1Why is sorting the array before summing elements > K suboptimal?

Sorting adds O(n log n) overhead, which is unnecessary because the relative order doesn't matter for the sum. A linear scan achieves the same result in O(n) time.

Q2How would you handle integer overflow when summing large values?

Use a wider integer type (e.g., 64-bit long) or language-specific big integer libraries. Additionally, check for overflow during accumulation if the language doesn't handle it automatically.

Q3Can this algorithm be parallelized, and what would be the trade‑off?

Yes, by partitioning the array and summing each chunk concurrently, then combining partial sums. The trade‑off is added synchronization and memory overhead, which may not pay off for modest input sizes.

Examples

Example 1

Input

[6, 7, 8, 9, 10], 5

Output

10

Explanation: Step-by-step: Given the input array [6, 7, 8, 9, 10] and K = 5, we iterate through the array and sum up all elements greater than 5. The elements 6, 7, 8, 9, and 10 are greater than 5, so we add them up to get the final output of 10.

Example 2

Input

[1, 2, 3, 4, 5], 5

Output

0

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and K = 5, we iterate through the array and sum up all elements greater than 5. However, there are no elements greater than 5 in the array, so we return 0 as the final output.

Constraints

  • 1 <= N <= 10^5
  • -10^4 <= metrics[i] <= 10^4
  • 1 <= K <= N

Optimal Approach & Strategy

The optimal solution scans the array once, compares each element to K, and adds qualifying values to a running sum, achieving O(n) time and O(1) space.

Brute Force Approach

A brute-force method might sort the array then sum from the first element greater than K, costing O(n log n) time.

Verified Code Solutions

JavaScript Solution
Time: O(n)
function solution(nums, k) {
      let sum = 0;
      for (let num of nums) {
         if (typeof num === 'number' && num > k) {
            sum += num;
         }
      }
      return sum;
   }

Asked in Top Tech Interviews

GoogleAmazonMicrosoft

Solve in Interative Editor

Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.