Sensor Packet Optimizer 2 — Problem Statement & Solution Guide
Problem Description
In a high-frequency telemetry pipeline, sensor nodes emit integer-encoded signal strengths. The network controller must filter this stream to isolate high-priority data. Given an array signals representing the raw metric values and an integer threshold, identify all elements that are strictly greater than threshold. Compute the cumulative sum of these qualifying values. If no elements exceed the threshold, return 0. The solution must efficiently traverse the data structure to aggregate the relevant metrics without unnecessary overhead.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Packet Optimizer 2"
WHY DOES IT MATTER?
The single‑pass filtering pattern is essential in streaming and real‑time analytics because it guarantees linear time complexity and minimal memory usage, which are the bottlenecks in high‑frequency data pipelines.
OPTIMIZATION CHALLENGE
The key insight is that the threshold comparison can be performed in constant time per element, and the sum can be accumulated without storing the qualifying elements, thus avoiding any additional data structures.
REAL-WORLD CONNECTION
Consider a network router that must drop or forward packets based on priority flags. The router performs a single comparison per packet and updates counters on the fly, mirroring the linear scan pattern used here.
When explaining this pattern in an interview, emphasize the importance of deterministic O(n) performance and the avoidance of branching where possible, perhaps by using bitwise masks to showcase low‑level optimization.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The optimal solution for filtering and summing elements greater than a threshold is a single linear scan of the array. In each iteration we compare the current element to the threshold and, if it satisfies the condition, add it to an accumulator. This approach runs in O(n) time and O(1) auxiliary space, making it suitable for high‑frequency telemetry streams where latency and memory footprint are critical.
A naive approach might involve repeatedly filtering the array or using nested loops to recompute partial sums, leading to O(n^2) time complexity. Such quadratic solutions quickly become infeasible as the number of sensor packets grows into the millions, causing unacceptable delays in real‑time systems. By contrast, the linear scan guarantees that each packet is processed exactly once, ensuring deterministic performance regardless of input size.
Bit manipulation can be leveraged in advanced implementations to reduce branching overhead—for example, using a mask derived from the comparison result to conditionally add values without an explicit if‑statement. However, the core algorithm remains a simple comparison and accumulation, and the primary optimization is the single pass over the data.
Interview Questions on This Problem
Q1How would you modify this algorithm to handle a stream of sensor data that arrives in real time, ensuring the sum is updated incrementally?
You would maintain a running sum and update it as each new packet arrives: if the packet value exceeds the threshold, add it to the sum; otherwise, ignore it. This allows constant‑time updates per packet and keeps memory usage minimal.
Q2A fintech platform needs to compute the sum of all transaction amounts above a dynamic threshold that changes every minute. What data structure would you use to support efficient updates and queries?
A segment tree or binary indexed tree (Fenwick tree) can be used to maintain prefix sums and support range queries. When the threshold changes, you can perform a binary search on the sorted transaction list to find the new cutoff and adjust the sum accordingly, achieving O(log n) update and query times.
Q3During a high‑growth startup interview, you are asked to explain why a single pass algorithm is preferable over a two‑pass approach for this problem. What points would you highlight?
A single pass reduces the number of memory accesses and eliminates the need to store intermediate results, which is critical in low‑latency environments. It also simplifies the code, reduces the risk of bugs, and ensures that the algorithm scales linearly with input size, meeting the performance expectations of fast‑moving startups.
Examples
Input
signals = [12, 4, 15, 8, 20], threshold = 10
Output
47
Explanation: Traverse the array: 12 > 10 (add 12), 4 <= 10 (skip), 15 > 10 (add 15), 8 <= 10 (skip), 20 > 10 (add 20). Sum = 12 + 15 + 20 = 47.
Input
signals = [5, 5, 5], threshold = 5
Output
0
Explanation: All elements are equal to the threshold. Since the condition requires strictly greater than, no elements qualify. Sum remains 0.
Input
signals = [-3, 0, 7, -1, 12], threshold = 0
Output
19
Explanation: Check each value: -3 <= 0 (skip), 0 <= 0 (skip), 7 > 0 (add 7), -1 <= 0 (skip), 12 > 0 (add 12). Sum = 7 + 12 = 19.
Constraints
- 1 <= signals.length <= 10^5
- -10^9 <= signals[i] <= 10^9
- -10^9 <= threshold <= 10^9
Optimal Approach & Strategy
The optimal method scans the array once, compares each element to the threshold, and accumulates the sum in a single variable. This yields linear time and constant extra space.
Brute Force Approach
A naive solution might repeatedly filter the array or use nested loops to recompute partial sums, resulting in quadratic time complexity. This approach is impractical for large datasets.
Verified Code Solutions
function solution(nums, K) { let sum = 0; for (let num of nums) { if (num > K) sum += num; } return sum; }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) { let sum = 0; for (let num of nums) { if (num > K) sum += num; } return sum; }Asked in Top Tech Interviews
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