Sensor Cluster Evaluator 41 — Problem Statement & Solution Guide

StackMediumMonotonic Stack

TimeO(N log K)

SpaceO(K)

Problem Description

A distributed sensor network transmits a sequence of integer readings representing signal strengths. The central processing unit must identify the K most significant signals to prioritize data aggregation. Given an array of integers readings and an integer K, compute the sum of the K largest elements in the array.

The input consists of a one-dimensional array readings where each element represents a distinct sensor's signal strength at a specific timestamp. The integer K specifies the count of top-tier signals to be aggregated. The output is a single integer representing the cumulative sum of these K largest values.

Note that the array may contain negative values, duplicates, and zero. The selection of the K largest elements is based on their numerical magnitude, regardless of their position in the array. If the array contains fewer than K elements, the problem guarantees that K will always be less than or equal to the length of the array.

DSA Pattern Breakdown

"Sensor Cluster Evaluator 41"

medium

WHY DOES IT MATTER?

This pattern is essential for 'Top-K' problems, which are ubiquitous in data analytics, recommendation systems, and monitoring dashboards. Mastering this teaches candidates how to avoid over-engineering (full sorts) and under-engineering (linear scans for each element), striking the right balance for partial ordering.

OPTIMIZATION CHALLENGE

The key insight is that we do not need a fully sorted array, only the boundary between the top K and the rest. By maintaining a Min-Heap of size K, we ensure that the smallest element in the heap is the K-th largest in the array. Any new element larger than this minimum replaces it, keeping the heap size constant and the sum updatable in O(1) or O(log K) time.

REAL-WORLD CONNECTION

In distributed sensor networks, the central unit cannot afford to sort millions of incoming packets. Instead, it maintains a 'window' of the top K strongest signals. This is analogous to a streaming window in Apache Kafka or a priority queue in a task scheduler that only cares about the highest priority tasks, not the entire queue order.

During the interview, explicitly state that you are choosing a Min-Heap over a Max-Heap because we want to evict the smallest of the 'top K' candidates. This demonstrates a clear understanding of the data structure's role in maintaining the invariant that the heap contains exactly the K largest elements seen so far.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log K)

💾 Space: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. The naive approach involves sorting the entire array in descending order and summing the first K elements. While straightforward, this incurs a time complexity of O(N log N), which is inefficient when K is significantly smaller than N. For large-scale sensor networks where N can be in the millions and K is a small constant (e.g., top 10 signals), the overhead of a full sort is unnecessary and computationally wasteful.

Interview Questions on This Problem

Q1How would you optimize the solution if K is very close to N (e.g., K = N - 1)?

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 reduces the problem size to min(K, N-K), allowing the use of a min-heap of size min(K, N-K) or a quickselect variant, thereby optimizing the constant factors and heap operations.

Q2What is the trade-off between using a Min-Heap and Quickselect for this problem?

A Min-Heap guarantees O(N log K) time complexity and is stable, making it predictable for real-time systems. Quickselect offers an average case of O(N) but has a worst-case of O(N^2). In production environments where latency spikes are unacceptable, the Min-Heap is often preferred for its consistent performance, whereas Quickselect might be chosen in offline batch processing where average-case speed is paramount.

Q3How would you handle duplicate values in the sensor readings when selecting the top K?

The algorithm naturally handles duplicates because it operates on values, not indices. If multiple sensors report the same signal strength, they are treated as distinct elements. The heap or selection algorithm will include all instances of a value if it falls within the top K threshold, ensuring the sum reflects the total signal strength of all significant sensors.

Examples

Example 1

Input

readings = [3, 1, 4, 1, 5, 9, 2, 6], K = 3

Output

Explanation: The array is [3, 1, 4, 1, 5, 9, 2, 6]. Sorting the elements in descending order yields [9, 6, 5, 4, 3, 2, 1, 1]. The top 3 elements are 9, 6, and 5. Their sum is 9 + 6 + 5 = 20.

Example 2

Input

readings = [-5, -1, -3, -2, -4], K = 2

Output

-3

Explanation: The array is [-5, -1, -3, -2, -4]. Sorting the elements in descending order yields [-1, -2, -3, -4, -5]. The top 2 elements are -1 and -2. Their sum is -1 + (-2) = -3.

Example 3

Input

readings = [10, 10, 10, 10], K = 4

Output

Explanation: The array is [10, 10, 10, 10]. All elements are identical. The top 4 elements are 10, 10, 10, and 10. Their sum is 10 + 10 + 10 + 10 = 40.

Example 4

Input

readings = [0, -1, 2, -3, 4], K = 1

Output

Explanation: The array is [0, -1, 2, -3, 4]. Sorting the elements in descending order yields [4, 2, 0, -1, -3]. The top 1 element is 4. The sum is 4.

Constraints

1 <= readings.length <= 10^5
-10^9 <= readings[i] <= 10^9
1 <= K <= readings.length

Optimal Approach & Strategy

Use a Min-Heap of size K to maintain the K largest elements encountered so far. Iterate through the array, replacing the smallest element in the heap if the current element is larger, then sum the heap contents at the end.

Brute Force Approach

Sort the entire array in descending order and sum the first K elements. This approach is simple but inefficient for large arrays as it requires O(N log N) time complexity regardless of the value of K.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log 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;
}

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];
       }
       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];
       }
       return sum;
   }
}

def solution(nums, k):
   nums.sort(reverse=True)
   sum = 0
   for i in range(k):
       sum += nums[i]
   return sum

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;
}

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Sensor Cluster Evaluator 41 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Stack

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Sensor Cluster Evaluator 41 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor