Sensor Cluster Optimizer 25 — Problem Statement & Solution Guide
Problem Description
Sensor Cluster Optimizer 25
You are given an array of integers, nums, and an integer K. Your task is to compute the sum of the K largest values present in nums. The array may contain negative numbers, zeros, and positive numbers, and the elements are not guaranteed to be sorted.
Input format: The first line contains two space‑separated integers, N and K, where N is the length of the array and K is the number of top elements to sum. The second line contains N space‑separated integers representing the elements of nums.
Output format: Output a single integer – the sum of the K largest elements in the array.
The problem requires an efficient solution that can handle large input sizes within reasonable time limits.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Optimizer 25"
WHY DOES IT MATTER?
Selecting top K elements is a foundational pattern in data processing, ranking systems, and resource allocation problems. Mastering it equips engineers to handle large stream processing with minimal memory overhead.
OPTIMIZATION CHALLENGE
The main challenge lies in eliminating full O(N log N) sorting in favor of O(N log K) heap tracking or average O(N) Quickselect selection.
REAL-WORLD CONNECTION
In distributed sensor networks, telemetry clusters frequently aggregate the top K strongest signals or highest power consumption readings for localized diagnostic analysis.
When streams of incoming data are non-stationary, using a bounded Min-Heap allows continuous maintenance of the top K elements in O(log K) per insertion without reprocessing past entries.
COMPLEXITY AT A GLANCE
O(N log K)O(K)Core Theory — Why This Approach?
To find the sum of the K largest elements in an unsorted array containing negative, zero, and positive values, we need to efficiently extract the top K elements without performing unnecessary operations on the rest of the collection. A naive strategy of sorting the entire array takes O(N log N) time, which becomes inefficient when N is very large and K is significantly smaller than N. Dynamic decision-making and optimal selection paradigms allow us to partition or maintain bounded state buffers to focus strictly on the target elements.
Interview Questions on This Problem
Q1How does using a Min-Heap of size K compare to sorting the entire array for finding the sum of K largest elements?
Sorting the full array takes O(N log N) time and O(1) or O(N) extra space depending on the sorting algorithm. Using a Min-Heap of size K reduces the time complexity to O(N log K) and bounds additional space to O(K), which is significantly faster when K is much smaller than N.
Q2Can Quickselect be used to solve this problem in average linear time?
Yes, Quickselect can find the K-th largest element in average O(N) time by repeatedly partitioning the array around a pivot. Once the array is partitioned such that the K largest elements are on one side, summing them up takes an additional O(K) time.
Q3How should negative numbers in the array be handled when selecting top K elements?
Negative numbers should be treated like any other real numbers since larger negative values (e.g., -2) are strictly greater than smaller ones (e.g., -10). The selection mechanism inherently prioritizes larger values regardless of their sign.
Examples
Input
5 3 1 3 5 7 9
Output
21
Explanation: The array sorted in descending order is [9, 7, 5, 3, 1]. The top 3 elements are 9, 7, and 5. Their sum is 9 + 7 + 5 = 21.
Input
6 2 -5 -2 -3 -1 -4 -6
Output
-3
Explanation: Sorting descending gives [-1, -2, -3, -4, -5, -6]. The two largest are -1 and -2. Their sum is -1 + (-2) = -3.
Input
4 4 10 20 30 40
Output
100
Explanation: All four elements are required. Summing them yields 10 + 20 + 30 + 40 = 100.
Input
7 5 0 5 -1 3 2 4 1
Output
15
Explanation: Descending order: [5, 4, 3, 2, 1, 0, -1]. The top 5 are 5, 4, 3, 2, and 1. Their sum is 5 + 4 + 3 + 2 + 1 = 15.
Constraints
- 1 <= N <= 100000
- 1 <= K <= N
- -1000000000 <= nums[i] <= 1000000000
- The result fits within a 64‑bit signed integer
Optimal Approach & Strategy
Maintain a Min-Heap of size K while iterating through the array, popping the smallest element whenever the size exceeds K. Finally, sum all elements remaining in the heap in O(N log K) time and O(K) space.
Brute Force Approach
Sort the entire array in descending order using standard sorting algorithms and then sum the first K elements. This takes O(N log N) time and up to O(N) auxiliary space.
Verified Code Solutions
function solution(nums, k) {
// Sort the array in descending order
nums.sort((a, b) => b - a);
// Handle the case where k is greater than the length of the array
k = Math.min(k, nums.length);
// Sum the first k elements in the sorted array
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
// Sort the array in descending order
sort(nums.begin(), nums.end(), greater<int>());
// Handle the case where k is greater than the length of the array
k = min(k, (int)nums.size());
// Sum the first k elements in the sorted array
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
// Sort the array in descending order
Arrays.sort(nums);
// Handle the case where k is greater than the length of the array
k = Math.min(k, nums.length);
// Sum the first k elements in the sorted array
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
# Sort the array in descending order
nums.sort(reverse=True)
# Handle the case where k is greater than the length of the array
k = min(k, len(nums))
# Sum the first k elements in the sorted array
return sum(nums[:k])function solution(nums, k) {
// Sort the array in descending order
nums.sort((a, b) => b - a);
// Handle the case where k is greater than the length of the array
k = Math.min(k, nums.length);
// Sum the first k elements in the sorted array
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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