Sensor Cluster Evaluator 6 — Problem Statement & Solution Guide
Problem Description
In a distributed sensor network, each node reports a metric value representing its current load. You are tasked with evaluating a specific cluster of sensors to determine the aggregate load of the most critical nodes. Given an array metrics of integers and an integer K, identify the sum of the K largest elements in the array, subject to the condition that each selected element must be strictly greater than K.
If fewer than K elements in the array satisfy the condition metrics[i] > K, return 0. If no elements satisfy the condition, return 0. The selection process must prioritize the largest values among those that meet the threshold.
Your function should accept the array metrics and the integer K as inputs and return a single integer representing the calculated sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Evaluator 6"
WHY DOES IT MATTER?
Selecting the top K elements is a fundamental pattern in data‑intensive applications—think leaderboards, recommendation engines, or alert thresholds—where you need the most significant items without fully sorting the dataset.
OPTIMIZATION CHALLENGE
The key insight is to avoid sorting the entire collection by maintaining a bounded min‑heap that only stores the current K best candidates, thereby reducing the per‑element work to O(log K) instead of O(log N).
REAL-WORLD CONNECTION
In a distributed sensor network, each node periodically reports load metrics. A central controller may need to quickly identify the K most overloaded nodes to trigger load‑balancing actions, mirroring the K‑largest selection problem.
During an interview, start with the heap idea, but also mention the quick‑select (Hoare's selection) algorithm as an alternative O(N) average‑case solution, showing depth of knowledge.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
The task of finding the sum of the K largest elements in an unsorted 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., millions of sensor readings) the extra log‑factor becomes a bottleneck, especially in latency‑sensitive distributed systems.
A more optimal paradigm leverages a min‑heap (priority queue) of fixed capacity K. By iterating through the array once, we keep the K largest elements seen so far in the heap: if the heap size is below K we push the element; otherwise we compare the current element with the heap’s minimum and replace it when larger. This maintains the invariant that the heap always contains the K biggest values, and each insertion or replacement costs O(log K). Consequently the overall runtime drops to O(N log K) with only O(K) auxiliary space, which scales dramatically better when K ≪ N.
Interview Questions on This Problem
Q1How would you compute the sum of the K largest numbers in a stream of integers where the total count is unknown beforehand?
Maintain a min‑heap of size K while reading the stream. For each incoming number, if the heap has fewer than K elements push it; otherwise, if the number exceeds the heap’s root, pop the root and push the new number. After the stream ends, sum the heap’s contents. This yields O(N log K) time and O(K) space.
Q2Can you modify the solution to also return the indices of the K largest elements in the original array?
Store pairs of (value, index) in the min‑heap instead of just values. The heap ordering uses the value component, preserving the index for later retrieval. After processing, extract the indices from the heap and optionally sort them if a specific order is required.
Q3What trade‑offs would you consider when K is close to N, say K = N/2, versus when K is very small?
When K is close to N, the overhead of maintaining a heap of size K approaches that of sorting; in such cases a full sort (O(N log N)) may be simpler and have comparable performance. Conversely, when K is tiny, the heap approach shines because log K is minimal, giving near‑linear performance. Choosing between heap and sort depends on the relative magnitude of K and N as well as constant‑factor considerations.
Examples
Input
metrics = [10, 20, 30, 40, 50], K = 3
Output
120
Explanation: Step 1: Identify elements strictly greater than K (3). All elements [10, 20, 30, 40, 50] are greater than 3. Step 2: Select the K (3) largest elements from this set. The sorted descending order is [50, 40, 30, 20, 10]. The top 3 are 50, 40, and 30. Step 3: Sum these values: 50 + 40 + 30 = 120. Step 4: Return 120.
Input
metrics = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Step 1: Identify elements strictly greater than K (10). No elements in [1, 2, 3, 4, 5] are greater than 10. Step 2: Since the count of valid elements is 0, which is less than K (10), the condition fails. Step 3: Return 0.
Input
metrics = [15, 5, 25, 10, 35], K = 2
Output
60
Explanation: Step 1: Identify elements strictly greater than K (2). All elements [15, 5, 25, 10, 35] are greater than 2. Step 2: Select the K (2) largest elements. Sorted descending: [35, 25, 15, 10, 5]. The top 2 are 35 and 25. Step 3: Sum these values: 35 + 25 = 60. Step 4: Return 60.
Input
metrics = [100, 200, 50, 150], K = 150
Output
0
Explanation: Step 1: Identify elements strictly greater than K (150). Only 200 is greater than 150. The valid set is [200]. Step 2: The count of valid elements is 1. Since 1 < K (150), we cannot select K elements. Step 3: Return 0.
Constraints
- 1 <= metrics.length <= 10^5
- 1 <= K <= 10^5
- 1 <= metrics[i] <= 10^9
Optimal Approach & Strategy
Use a min‑heap of size K to keep track of the K largest values while iterating once through the array.
Brute Force Approach
Sort the entire array in descending order and sum the first K elements.
Verified Code Solutions
function solution(nums, k) {
let filtered = nums.filter(num => num > k);
return filtered.reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
}def solution(nums, k):
filtered = [num for num in nums if num > k]
return sum(filtered)function solution(nums, k) {
let filtered = nums.filter(num => num > k);
return filtered.reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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