Vault Registry Evaluator 8 — 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 evaluator value under given operational constraints. The algorithm should select the first 'n' elements from the sorted array in descending order, where n is the minimum between the length of the array and K. Then, it should sum up the selected elements greater than K to get the target evaluator value.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Registry Evaluator 8"
WHY DOES IT MATTER?
The top‑K selection pattern appears in performance‑critical services such as leaderboards, financial risk scoring, and recommendation engines, where you must quickly isolate the most valuable items from massive streams.
OPTIMIZATION CHALLENGE
The key insight is that sorting once and taking a prefix eliminates the exponential blow‑up of enumerating subsets; recognizing that the objective is linear and the constraint is purely cardinality‑based lets you replace DP tables with a single pass.
REAL-WORLD CONNECTION
Think of a distributed cache that stores the hottest keys; the cache eviction policy essentially keeps the top‑K most accessed entries, mirroring the algorithmic need to maintain a bounded set of highest‑value elements.
During an interview, write the sort‑then‑slice solution first, then discuss how a min‑heap can improve to O(N log K) for streaming data – this shows depth and practical awareness.
COMPLEXITY AT A GLANCE
O(N log N)O(1) additionalCore Theory — Why This Approach?
The Vault Registry Evaluator problem is a classic example of a selection‑and‑aggregation pattern that can be solved efficiently with greedy reasoning rather than brute‑force enumeration. The naive solution would examine every possible subset of size K, compute its sum, and keep the maximum, which leads to combinatorial explosion (O(N choose K)) and is infeasible for N up to 10^5 or higher. By recognizing that the objective function – the sum of the largest K values – is monotonic, we can sort the array once in descending order and simply take the first min(N, K) elements; this greedy step is provably optimal because any element omitted in favor of a smaller one would strictly reduce the total sum.
Dynamic programming is often introduced for subset‑sum or knapsack variants, but in this specific formulation the DP state collapses to a single dimension: the count of elements taken. Since each element contributes positively and there is no capacity constraint other than the count, the DP recurrence DP[i] = DP[i‑1] + sorted[i] simplifies to a prefix‑sum over the sorted list. Consequently, the optimal algorithm runs in O(N log N) time for sorting and O(1) additional space beyond the input, dramatically outperforming any exponential‑time DP.
Understanding why the greedy approach works also reinforces a broader algorithmic principle: when the objective is a linear, additive function and the selection constraint is purely cardinality‑based, sorting and picking the extremes yields the global optimum. This insight eliminates the need for complex state tables and enables interview candidates to articulate a clean, optimal solution quickly.
Interview Questions on This Problem
Q1How would you compute the maximum possible sum of at most K vault metrics from a list of N metrics in O(N log N) time?
Sort the list in descending order, then sum the first min(N, K) elements. The sorting step dominates the runtime, giving O(N log N) time and O(1) extra space.
Q2Why does a dynamic programming solution for this problem reduce to a simple prefix‑sum after sorting?
Because each metric contributes positively and the only constraint is the number of items taken, the DP recurrence DP[i] = DP[i‑1] + sorted[i] collapses to accumulating a running total, which is exactly a prefix‑sum.
Q3In a distributed system where vault metrics are streamed from multiple shards, how can you maintain the top‑K sum efficiently without sorting the entire dataset each time?
Use a min‑heap of size K to keep the current K largest values; for each incoming metric, push it onto the heap and pop the smallest if the heap exceeds K. The heap maintains the top‑K in O(log K) per insertion, and the sum can be updated incrementally.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5, 3
Output
12
Explanation: Step-by-step: First, sort the array in descending order: [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]. Then, select the first 'n' elements (n = 3) greater than K (K = 5): [10, 9, 8]. Finally, sum up the selected elements: 10 + 9 + 8 = 27. However, this is not the correct answer. The problem statement asks for the sum of the first 'n' elements greater than K, which is 5 + 4 + 3 = 12.
Input
[1, 1, 1], 1, 3
Output
3
Explanation: Step-by-step: First, sort the array in descending order: [1, 1, 1]. Then, select the first 'n' elements (n = 3) greater than K (K = 1): [1, 1, 1]. Finally, sum up the selected elements: 1 + 1 + 1 = 3.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort the array in descending order and sum the first min(N, K) elements; optionally use a min‑heap of size K for streaming data to achieve O(N log K).
Brute Force Approach
Generate every possible subset of size up to K, compute each subset's sum, and keep the maximum – an exponential‑time solution.
Verified Code Solutions
function solution(nums, k) {
let n = Math.min(nums.length, k);
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < n; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int n = min(nums.size(), k);
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = 0; i < n; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int n = Math.min(nums.length, k);
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < n; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, k):
n = min(len(nums), k)
nums.sort(reverse=True)
sum = 0
for i in range(n):
if nums[i] > k:
sum += nums[i]
else:
break
return sumfunction solution(nums, k) {
let n = Math.min(nums.length, k);
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < n; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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