Protocol Tome Validator 22 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and tome metrics, construct an optimal algorithm to evaluate and compute the target validator value under given operational constraints. The target validator value is the sum of the K largest values in the sequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Validator 22"
WHY DOES IT MATTER?
Selecting top‑K elements is a fundamental building block for ranking, recommendation, and resource allocation systems.
OPTIMIZATION CHALLENGE
The key is to keep the data structure bounded at K to cut both time and memory from O(N) to O(K).
REAL-WORLD CONNECTION
Search engines keep a min‑heap of the best results while scanning billions of documents.
Initialize the heap with the first K items, then reuse the same structure to avoid reallocations during the scan.
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 a list is a classic selection problem that can be solved efficiently with a heap. A min‑heap of size K maintains the current K biggest values; each new element larger than the heap root replaces it, guaranteeing that after a single pass the heap contains exactly the K largest numbers.
Naïve approaches—sorting the entire array (O(N log N)) or scanning all subsets (exponential)—become prohibitive for large N (up to 10^6 or more). By limiting heap size to K, we reduce the per‑element cost to O(log K), yielding an overall O(N log K) solution that scales gracefully even when K is much smaller than N.
Interview Questions on This Problem
Q1Why is a min‑heap of size K preferred over sorting the whole array for this problem?
A min‑heap only stores K elements, so each insertion or replacement costs O(log K) instead of O(log N). This reduces total time to O(N log K), which is faster when K ≪ N.
Q2How would you handle the case when K is larger than the array length?
Clamp K to the array size before processing, effectively summing all elements. This avoids out‑of‑bounds errors and ensures a correct result.
Q3What is the impact of using a max‑heap and extracting K times versus the min‑heap approach?
A max‑heap requires building a heap of size N (O(N)) and then K pop operations (O(K log N)), which is slower when K is small. The min‑heap method avoids the extra O(N) space and reduces per‑operation cost.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
12
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], K = 3, we first sort the array in descending order to get [5, 4, 3, 2, 1]. Then, we sum the first K elements, which are 5, 4, and 3, giving us a total of 12.
Input
[10, 10, 10, 10, 10], 3
Output
30
Explanation: Step-by-step: with input [10, 10, 10, 10, 10], K = 3, we first sum all the elements in the array, which is 50. Since the sum of the K largest values is less than the largest value in the array, we return the sum of all elements in the array, which is 50.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a min‑heap of size K while scanning the array once, achieving O(N log K) time and O(K) space.
Brute Force Approach
Sort the entire array and sum the last K elements, which costs O(N log N) time and O(N) space.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
if (sum < Math.max(...nums)) {
return nums.reduce((a, b) => a + b, 0);
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.rbegin(), nums.rend());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
if (sum < *max_element(nums.begin(), nums.end())) {
return accumulate(nums.begin(), nums.end(), 0);
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
if (sum < Arrays.stream(nums).max().getAsInt()) {
return Arrays.stream(nums).sum();
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
total = 0
for i in range(k):
total += nums[i]
if total < max(nums):
return sum(nums)
return totalfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
if (sum < Math.max(...nums)) {
return nums.reduce((a, b) => a + b, 0);
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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