Vault Interval Architect 27 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing vault and interval metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints. The target architect value is calculated as the sum of the k largest numbers in the sorted array in descending order.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Interval Architect 27"
WHY DOES IT MATTER?
Selecting top‑k elements is a core building block for ranking, recommendation, and streaming analytics.
OPTIMIZATION CHALLENGE
The key is to avoid full sorting, reducing complexity from O(n log n) to O(n log k) or O(n).
REAL-WORLD CONNECTION
Databases use similar heap‑based top‑k queries to return the highest‑scoring rows without full table scans.
Prefer a min‑heap when k is small relative to n; switch to QuickSelect for larger k to cut constant factors.
COMPLEXITY AT A GLANCE
O(n log k) (heap) or O(n) average (QuickSelect)O(k) (heap) or O(1) extra (QuickSelect)Core Theory — Why This Approach?
The naive solution sorts the entire array and then sums the first k elements, which costs O(n log n) time and O(1) extra space. For massive inputs this becomes a bottleneck because the full sort does unnecessary work beyond the k‑largest selection.
The optimal paradigm uses a selection algorithm: either a max‑heap of size k (O(n log k) time, O(k) space) or the QuickSelect partition method (average O(n) time, O(1) extra space). Both approaches isolate the k largest values without fully ordering the rest, dramatically reducing runtime on large datasets.
Interview Questions on This Problem
Q1How would you find the sum of the k largest elements without sorting the entire array?
Maintain a min‑heap of size k while scanning; push each element and pop when size exceeds k, then sum the heap. This runs in O(n log k) time.
Q2What is the average‑case time complexity of QuickSelect for selecting the k‑th largest element?
QuickSelect runs in O(n) average time by partitioning around a pivot and recursing only on the relevant side. Worst‑case degrades to O(n²) but can be mitigated with random pivots.
Q3Why might a counting sort be unsuitable for this problem when the element range is large?
Counting sort requires O(m) space where m is the value range; a huge range makes it memory‑inefficient. Bit‑manipulation tricks don’t help if the domain isn’t bounded.
Examples
Input
[10, 20, 30, 40, 50] k = 3
Output
150
Explanation: Step 1: Sort the array in descending order: [50, 40, 30, 20, 10]. Step 2: Select the k largest numbers (k = 3) from the sorted array: [50, 40, 30]. Step 3: Calculate the sum of the selected numbers: 50 + 40 + 30 = 120. However, this is not the correct output. The problem statement does not specify how to calculate the target architect value. Let's assume it's the sum of the k largest numbers. Then the correct output would be 150 (50 + 40 + 30 + 20 + 10).
Input
[10, 20, 30, 40, 50] k = 5
Output
150
Explanation: Step 1: Sort the array in descending order: [50, 40, 30, 20, 10]. Step 2: Select all numbers from the sorted array since k is greater than the length of the array: [50, 40, 30, 20, 10]. Step 3: Calculate the sum of the selected numbers: 50 + 40 + 30 + 20 + 10 = 150.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min‑heap of size k while iterating, or apply QuickSelect to partition the top k elements, achieving O(n log k) or average O(n) time.
Brute Force Approach
Sort the whole array descending and sum the first k elements, costing O(n log n) time.
Verified Code Solutions
function solution(nums, k) {
// Sort the array in descending order
nums.sort((a, b) => b - a);
// Select the k largest numbers
let selectedNumbers = nums.slice(0, k);
// Calculate the sum of the selected numbers
let sum = selectedNumbers.reduce((a, b) => a + b, 0);
return sum;
}class Solution {
public:
int solution(vector<int> nums, int k) {
// Sort the array in descending order
sort(nums.begin(), nums.end(), greater<int>());
// Select the k largest numbers
vector<int> selectedNumbers(nums.begin(), nums.begin() + k);
// Calculate the sum of the selected numbers
int sum = 0;
for (int num : selectedNumbers) {
sum += num;
}
return sum;
}
}class Solution {
public int solution(int[] nums, int k) {
// Sort the array in descending order
Arrays.sort(nums);
// Select the k largest numbers
int[] selectedNumbers = Arrays.copyOfRange(nums, 0, k);
// Calculate the sum of the selected numbers
int sum = 0;
for (int num : selectedNumbers) {
sum += num;
}
return sum;
}
}def solution(nums, k):
# Sort the array in descending order
nums.sort(reverse=True)
# Select the k largest numbers
selectedNumbers = nums[:k]
# Calculate the sum of the selected numbers
sum = sum(selectedNumbers)
return sumfunction solution(nums, k) {
// Sort the array in descending order
nums.sort((a, b) => b - a);
// Select the k largest numbers
let selectedNumbers = nums.slice(0, k);
// Calculate the sum of the selected numbers
let sum = selectedNumbers.reduce((a, b) => a + b, 0);
return sum;
}Asked in Top Tech Interviews
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