Tome Voyage Detector 31 — Problem Statement & Solution Guide
Problem Description
Given an array of integers and an integer K, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints. The target detector value is the sum of the K largest elements in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Detector 31"
WHY DOES IT MATTER?
Selecting the top K elements is a fundamental pattern in data‑driven systems—think of leaderboard generation, risk‑scoring top exposures, or caching the hottest items. Mastering this pattern lets engineers extract high‑value signals without incurring the cost of full sorting.
OPTIMIZATION CHALLENGE
The key insight is to maintain only the K most promising candidates at any time. By discarding smaller elements early via a bounded heap (or partitioning around a pivot), we avoid the O(N log N) overhead of ordering the entire dataset.
REAL-WORLD CONNECTION
In a distributed log‑processing pipeline, each node may keep a min‑heap of the K most frequent error codes. When aggregating across nodes, only the K‑largest counters need to be merged, dramatically reducing network traffic and storage.
During an interview, start with the heap solution because it’s easy to reason about correctness and complexity; then, if prompted, discuss QuickSelect and its in‑place nature to show depth of knowledge.
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 array is a classic selection problem. A naïve solution would sort the entire array, which costs O(N log N) time, and then sum the last K entries. While correct, this approach wastes work because we only need the top K values, not a full ordering of all N elements. For large N (e.g., N up to 10^7) and small K, sorting becomes a bottleneck both in time and memory, especially when the array resides in external storage or when the runtime budget is tight.
The optimal paradigm leverages a min‑heap (priority queue) of size K or the QuickSelect algorithm. By maintaining a min‑heap that stores the current K largest numbers, each new element is compared to the heap root in O(1) and potentially inserted in O(log K), yielding an overall O(N log K) time complexity with O(K) extra space. QuickSelect can achieve average O(N) time and O(1) extra space by partitioning the array around a pivot until the K‑th largest element is positioned, after which a linear scan sums the top K values. Both techniques dramatically reduce work compared to full sorting, making them suitable for high‑throughput or memory‑constrained environments.
Interview Questions on This Problem
Q1How would you compute the sum of the K largest numbers in an unsorted array when N is up to 10^6 and K is much smaller than N?
Use a min‑heap of size K. Iterate through the array, push the first K elements into the heap, then for each remaining element, if it is larger than the heap root replace the root (heapify). After processing, sum all elements in the heap. This runs in O(N log K) time and O(K) space.
Q2Explain the trade‑offs between using a min‑heap versus QuickSelect for this problem.
A min‑heap guarantees O(N log K) worst‑case time and uses O(K) extra space, which is predictable and easy to implement. QuickSelect offers average O(N) time and O(1) extra space but has O(N^2) worst‑case without randomization, and it mutates the input array. Choose a heap when K is small or when worst‑case guarantees matter; choose QuickSelect when memory is at a premium and average‑case performance is acceptable.
Q3What edge cases must you handle when K equals 0, K equals N, or the array contains duplicate values?
If K is 0, the answer is 0 regardless of array content. If K equals N, the sum is simply the total sum of the array, which can be computed in O(N) without extra structures. Duplicates are fine; the algorithm treats each element independently, so repeated values are counted as many times as they appear among the top K.
Examples
Input
[9, 8, 7, 6, 5, 4, 3, 2, 1]
Output
24
Explanation: Step-by-step: First, sort the array in descending order. Then, select the first K elements from the sorted array and sum them up. For the input [9, 8, 7, 6, 5, 4, 3, 2, 1], the sorted array is [9, 8, 7, 6, 5, 4, 3, 2, 1]. Selecting the first 3 elements (K = 3) gives us [9, 8, 7]. Summing them up gives us 24.
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90]
Output
150
Explanation: Step-by-step: First, sort the array in descending order. Then, select the first K elements from the sorted array and sum them up. For the input [10, 20, 30, 40, 50, 60, 70, 80, 90], the sorted array is [90, 80, 70, 60, 50, 40, 30, 20, 10]. Selecting the first 3 elements (K = 3) gives us [90, 80, 70]. Summing them up gives us 240.
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, inserting or replacing elements as needed, then sum the heap contents.
Brute Force Approach
Sort the entire array in descending order and sum the first K elements.
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];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = nums.size() - k; i < nums.size(); i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = nums.length - k; i < nums.length; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
return sum(nums[:k])function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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