Vault Registry Detector 40 — Problem Statement & Solution Guide
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"
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
O(n)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
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.
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
function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (typeof num === 'number' && 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 isinstance(num, (int, float)) and num > k:
sum += num
return sumfunction 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
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