Vault Registry Architect 16 — 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 architect value under given operational constraints. The target architect value is the sum of the K largest elements in the sequence. If K is greater than or equal to the length of the array, return the sum of all elements in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Registry Architect 16"
WHY DOES IT MATTER?
This pattern is essential for problems where you need to find the top K elements without fully sorting the array. It is a fundamental technique in competitive programming and system design, especially when dealing with large datasets where sorting is computationally expensive.
OPTIMIZATION CHALLENGE
The key insight is to use a min-heap of size K to keep track of the K largest elements seen so far. By only maintaining a heap of size K, we reduce the time complexity from O(N log N) to O(N log K), which is a significant improvement when K is small.
REAL-WORLD CONNECTION
In distributed systems, this pattern is used in load balancing to identify the top K most active nodes or in data analytics to find the top K most frequent queries. It allows for efficient real-time processing without the need to sort the entire dataset.
During an interview, always mention the trade-off between time and space complexity. If the interviewer asks for the most efficient solution, suggest the heap approach and explain why it is better than sorting when K is small. Also, consider edge cases like K >= N or K = 0.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The problem of finding the sum of the K largest elements in an unsorted array is a classic selection problem that can be solved using various paradigms, but the most efficient approach for large datasets involves the use of a min-heap (priority queue). The naive approach of sorting the entire array takes O(N log N) time, which is suboptimal when K is significantly smaller than N. By maintaining a min-heap of size K, we can process each element in the array in O(log K) time, resulting in an overall time complexity of O(N log K). This is superior to sorting when K << N, as it avoids the overhead of ordering elements that are not part of the top K.
Interview Questions on This Problem
Q1How would you optimize the solution if K is very close to N?
If K is close to N, it is more efficient to find the N-K smallest elements and subtract their sum from the total sum of the array. This can be done using a max-heap of size N-K, reducing the time complexity to O(N log(N-K)), which is better than O(N log K) when K is large.
Q2What are the space complexity implications of using a heap versus sorting?
Using a heap of size K requires O(K) additional space, whereas sorting in-place (if allowed) requires O(1) extra space but modifies the original array. If the array cannot be modified, sorting would require O(N) space for a copy, making the heap approach more space-efficient when K < N.
Q3How would you handle duplicate values in the array?
Duplicates are handled naturally by the heap approach. If multiple elements have the same value and are among the top K, they will all be included in the heap. The sum will correctly account for all duplicates, as the heap does not distinguish between identical values.
Examples
Input
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
Output
40
Explanation: Step-by-step: Given the array [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] and K = 5, we first sort the array in descending order. The 5 largest elements are 10, 9, 8, 7, and 6. We then return the sum of these elements, which is 40.
Input
[90, 80, 70, 60, 50, 40, 30, 20, 10]
Output
350
Explanation: Step-by-step: Given the array [90, 80, 70, 60, 50, 40, 30, 20, 10] and K = 5, we first sort the array in descending order. The 5 largest elements are 90, 80, 70, 60, and 50. We then return the sum of these elements, which is 350.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min-heap of size K to keep track of the K largest elements. Iterate through the array, adding elements to the heap and replacing the smallest element if the current element is larger. The time complexity is O(N log K), which is optimal for small K.
Brute Force Approach
Sort the array in descending order and sum the first K elements. This approach is simple but has a time complexity of O(N log N), which is inefficient for large arrays when K is small.
Verified Code Solutions
function solution(nums, k) {
if (k >= nums.length) return nums.reduce((a, b) => a + b, 0);
return nums.sort((a, b) => b - a).slice(0, k).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k >= nums.size()) return accumulate(nums.begin(), nums.end(), 0);
return accumulate(nums.rbegin(), nums.rbegin() + k, 0);
}
};class Solution {
public int solution(int[] nums, int k) {
if (k >= nums.length) return Arrays.stream(nums).sum();
return Arrays.stream(nums).sorted((a, b) -> b - a).limit(k).sum();
}
}def solution(nums, k):
if k >= len(nums):
return sum(nums)
return sum(sorted(nums, reverse=True)[:k])function solution(nums, k) {
if (k >= nums.length) return nums.reduce((a, b) => a + b, 0);
return nums.sort((a, b) => b - a).slice(0, k).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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.