Sensor Packet Analyzer 5 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and packet metrics, construct an optimal algorithm to evaluate and compute the target analyzer value under given operational constraints. The algorithm should sum the K largest elements in the sequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Packet Analyzer 5"
WHY DOES IT MATTER?
This pattern forms the foundation for top-K selection problems across data analytics, recommendation systems, and real-time monitoring. Mastering it enables engineers to efficiently process large datasets without full sorting, a critical skill for building scalable data pipelines and low-latency analytics engines.
OPTIMIZATION CHALLENGE
The key insight is recognizing that we only need to maintain K elements at any time. By using a min-heap instead of sorting, we reduce comparisons from O(n log n) to O(n log K) while keeping space complexity proportional to K rather than n.
REAL-WORLD CONNECTION
Network intrusion detection systems use this exact pattern to identify the most frequent source IPs in traffic logs. Financial trading platforms apply it to track top-performing assets in real-time market data streams, where processing speed directly impacts trading decisions.
Always verify heap size constraints before implementation. In interviews, explicitly mention why a min-heap is preferred over max-heap for this problem, and discuss how to handle edge cases like K=0 or K>n gracefully to demonstrate production-ready thinking.
COMPLEXITY AT A GLANCE
O(n log K)O(K)Core Theory — Why This Approach?
The problem of summing the K largest elements in a sequence is a classic application of greedy algorithms combined with priority queue optimization. Naive approaches that sort the entire array first incur O(n log n) time complexity, which becomes prohibitive for large-scale sensor data streams where n can exceed millions of elements. The optimal paradigm leverages a min-heap of fixed size K to maintain only the top K candidates encountered so far. By processing elements sequentially and evicting the smallest element when the heap exceeds capacity, we achieve O(n log K) time complexity while using O(K) auxiliary space. This approach is theoretically optimal because it avoids unnecessary comparisons between elements that cannot possibly belong to the top K subset, exploiting the mathematical property that the sum of K largest values depends only on their relative ordering within the selection window rather than global sequence arrangement.
Interview Questions on This Problem
Q1How would you modify this algorithm to handle streaming sensor data where elements arrive continuously and K changes dynamically?
Implement a dual-heap system with a min-heap for current top K elements and a max-heap for overflow elements. When K increases, transfer elements from the overflow heap until capacity matches new K. When K decreases, remove excess elements from the min-heap. Maintain running sum by adjusting for heap transfers, achieving O(log n) per update operation.
Q2What happens to the algorithm's performance when K approaches n/2 versus K=1? How would you optimize for these boundary cases?
When K≈n/2, the min-heap approach degrades to O(n log n) similar to sorting. For K=1, use a single pass tracking maximum element in O(n) time. Implement adaptive logic that switches to quickselect algorithm when K > n/3, maintaining O(n) average time complexity across all K values while preserving O(1) space for K=1 case.
Q3How would you handle duplicate sensor readings that might affect the sum calculation accuracy?
Maintain a frequency map alongside the min-heap to track duplicate counts. When evicting elements, decrement frequencies instead of removing duplicates. During sum calculation, multiply each unique value by its frequency. This preserves O(n log K) time complexity while ensuring accurate summation for datasets with high cardinality overlap.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and k = 5
Output
40
Explanation: Step-by-step: Sort the array in descending order to get [10, 9, 8, 7, 6, 5, 4, 3, 2, 1], then sum the 5 largest elements: 10 + 9 + 8 + 7 + 6 = 40
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and k = 3
Output
27
Explanation: Step-by-step: Sort the array in descending order to get [10, 9, 8, 7, 6, 5, 4, 3, 2, 1], then sum the 3 largest elements: 10 + 9 + 8 = 27
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 iterating through the array. For each element, push to heap and pop the smallest if size exceeds K. Sum the remaining heap elements. This achieves O(n log K) time and O(K) space complexity.
Brute Force Approach
Sort the entire array in descending order and sum the first K elements. This approach requires O(n log n) time for sorting and O(n) space to store the sorted array.
Verified Code Solutions
function solution(nums, k) {
if (k >= nums.length) return nums.reduce((a, b) => a + b, 0);
return nums.sort((a, b) => b - a).slice(0, k).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k >= nums.size()) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
sort(nums.rbegin(), nums.rend());
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) {
if (k >= nums.length) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
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):
if k >= len(nums):
return sum(nums)
return sum(sorted(nums, reverse=True)[:k])function solution(nums, k) {
if (k >= nums.length) return nums.reduce((a, b) => a + b, 0);
return nums.sort((a, b) => b - a).slice(0, k).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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