Sensor Checkpoint Architect 3 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing a distributed sensor network where each node reports a metric value. Given an array metrics of integers representing these readings and a threshold integer K, compute the total sum of all metric values that are greater than or equal to K. This sum represents the 'architect value' used for system calibration.
Your solution must process the input in a single pass to ensure efficiency, achieving a time complexity of O(n), where n is the length of the metrics array. The space complexity should be O(1) if we exclude the input storage.
The function should return the computed sum as an integer. If no metrics meet the threshold condition, the function should return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Checkpoint Architect 3"
WHY DOES IT MATTER?
The filter‑and‑aggregate pattern is fundamental because many real‑world analytics tasks involve selecting data that meets a threshold and then summarizing it. Mastery of this pattern enables efficient handling of large streams without unnecessary sorting or extra passes.
OPTIMIZATION CHALLENGE
Recognizing that the predicate (value >= K) is independent of element order eliminates the need for sorting or auxiliary data structures; the key insight is that a running total suffices, collapsing both filtering and summation into one loop.
REAL-WORLD CONNECTION
In a distributed sensor network, each node periodically sends its reading to a central controller. The controller must quickly compute the total of readings above a safety threshold to decide if corrective actions are needed, mirroring the single‑pass sum computation.
During an interview, write the loop that both checks the condition and updates the accumulator in the same line; this demonstrates concise thinking and reduces the chance of off‑by‑one or missed‑element bugs.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a classic linear aggregation over a collection, where we only include elements satisfying a predicate (value >= K). A naive solution might attempt to sort the array or use nested loops to compare each element with every other, which inflates the time complexity to O(N log N) or O(N^2) and is unnecessary because the predicate is independent of element ordering. The optimal paradigm leverages a single-pass scan, maintaining a running total that only adds qualifying values, achieving O(N) time with O(1) auxiliary space. This approach exemplifies the "filter‑and‑aggregate" pattern, a staple in algorithmic design for problems that require selective summation or counting.
When dealing with large inputs—potentially millions of sensor readings—the single-pass method ensures the program stays within memory limits and executes within tight time constraints typical of real‑time systems. Moreover, using a 64‑bit accumulator (e.g., long long) prevents overflow when the sum of qualifying metrics exceeds the 32‑bit integer range, a subtle but critical detail in production‑grade code.
Interview Questions on This Problem
Q1How would you modify the solution if the requirement changed to finding the sum of the top M metric values instead of all values >= K?
Use a min‑heap of size M to keep the largest M elements while scanning the array. For each element, push it onto the heap; if the heap size exceeds M, pop the smallest. After the scan, sum the heap contents. This runs in O(N log M) time and O(M) space.
Q2What considerations would you make for handling integer overflow in this problem, especially in languages without built‑in big integer support?
Cast the accumulator to a 64‑bit type (e.g., long long in C++/Java) before adding any element. Additionally, check for potential overflow by comparing against the maximum value of the type if the problem constraints allow values near the limit.
Q3Explain how you could parallelize the computation for a massive distributed sensor dataset.
Partition the array across multiple workers, each computing a local sum of values >= K. Then perform a reduction (e.g., MPI_Reduce or a map‑reduce combine step) to aggregate the local sums into the final result. This yields O(N/p) time per worker with O(p) communication overhead, where p is the number of workers.
Examples
Input
metrics = [10, 20, 30, 40, 50], K = 25
Output
120
Explanation: Iterate through the array: 1. 10 < 25, skip. 2. 20 < 25, skip. 3. 30 >= 25, add 30 to sum (sum = 30). 4. 40 >= 25, add 40 to sum (sum = 70). 5. 50 >= 25, add 50 to sum (sum = 120). Final result is 120.
Input
metrics = [5, 15, 25, 35, 45], K = 50
Output
0
Explanation: Iterate through the array: 1. 5 < 50, skip. 2. 15 < 50, skip. 3. 25 < 50, skip. 4. 35 < 50, skip. 5. 45 < 50, skip. No elements meet the condition, so the sum remains 0.
Input
metrics = [100, 90, 80, 70, 60], K = 75
Output
270
Explanation: Iterate through the array: 1. 100 >= 75, add 100 to sum (sum = 100). 2. 90 >= 75, add 90 to sum (sum = 190). 3. 80 >= 75, add 80 to sum (sum = 270). 4. 70 < 75, skip. 5. 60 < 75, skip. Final result is 270.
Input
metrics = [0, -10, 10, -20, 20], K = 0
Output
30
Explanation: Iterate through the array: 1. 0 >= 0, add 0 to sum (sum = 0). 2. -10 < 0, skip. 3. 10 >= 0, add 10 to sum (sum = 10). 4. -20 < 0, skip. 5. 20 >= 0, add 20 to sum (sum = 30). Final result is 30.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Iterate once, adding each element that satisfies value >= K to a running total, achieving O(N) time and O(1) space.
Brute Force Approach
Sort the array and then sum elements from the first index that meets the threshold, which costs O(N log N) time.
Verified Code Solutions
function solution(nums, K) { return nums.filter(num => num >= K).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): return sum(num for num in nums if num >= K)function solution(nums, K) { return nums.filter(num => num >= K).reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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