Protocol Pipeline Partition 2 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a sequence of integer metrics derived from a distributed protocol pipeline. Given an array metrics and an integer threshold K, your objective is to compute the aggregate value of all metrics that satisfy the condition metrics[i] >= K.
This problem requires you to iterate through the data stream and selectively accumulate values based on the specified lower bound. The solution should efficiently handle large datasets while maintaining strict adherence to the filtering criteria. The final result represents the total weight of qualifying pipeline segments.
Input: An array of integers metrics representing the data points and an integer K representing the minimum threshold.
Output: A single integer representing the sum of all elements in metrics that are greater than or equal to K. If no elements meet the criteria, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Pipeline Partition 2"
WHY DOES IT MATTER?
Linear aggregation is a foundational pattern in data processing; mastering it ensures candidates can solve a wide range of problems efficiently without overengineering.
OPTIMIZATION CHALLENGE
The key insight is that each element’s contribution is independent; thus, no sorting or grouping is required, and a single accumulator suffices.
REAL-WORLD CONNECTION
Think of a monitoring system that aggregates error counts above a threshold across distributed services—each service reports metrics independently, and the aggregator simply sums those that exceed the alert level.
When explaining this in an interview, emphasize the O(n) time and O(1) space, and note that the algorithm is inherently parallelizable across multiple cores if needed.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a simple linear scan: iterate over each metric, compare it to the threshold K, and accumulate the value if it meets or exceeds K. A naive approach might sort the array first or use nested loops to filter and sum, which would add unnecessary O(n log n) or O(n^2) overhead and waste memory for temporary storage. By recognizing that the condition is independent for each element, we can apply a single-pass, constant‑space algorithm that achieves optimal time complexity of O(n) and space complexity of O(1). This pattern is a classic example of a linear aggregation problem where the only state needed is the running sum, making it both efficient and straightforward to implement.
Interview Questions on This Problem
Q1How would you handle this problem if the metrics array were extremely large and streamed in real time, potentially exceeding available memory?
I would process the stream in a single pass, maintaining only a running total. If the stream is truly infinite, I would also consider using a sliding window or a bounded buffer if only recent metrics are relevant, but for the sum of all qualifying metrics, a single accumulator suffices.
Q2In a fintech application, why is it important to avoid sorting or additional data structures for this type of threshold aggregation?
Sorting or extra structures introduce O(n log n) or O(n) space overhead, which can be costly in latency‑sensitive environments like real‑time trading. A linear scan keeps the algorithm deterministic and low‑latency, ensuring that the system can meet strict SLA requirements.
Q3What would be the impact on performance if you mistakenly used a recursive approach to sum the qualifying metrics?
A recursive solution would add O(n) call stack overhead and risk stack overflow for large inputs. It also obscures the linear nature of the problem and can degrade cache locality, leading to slower execution compared to an iterative loop.
Examples
Input
metrics = [12, 5, 8, 15, 3], K = 10
Output
27
Explanation: Iterate through the array: 1. 12 >= 10: Add 12 to sum (sum = 12). 2. 5 < 10: Skip. 3. 8 < 10: Skip. 4. 15 >= 10: Add 15 to sum (sum = 27). 5. 3 < 10: Skip. Final sum is 27.
Input
metrics = [1, 2, 3, 4, 5], K = 6
Output
0
Explanation: Iterate through the array: 1. 1 < 6: Skip. 2. 2 < 6: Skip. 3. 3 < 6: Skip. 4. 4 < 6: Skip. 5. 5 < 6: Skip. No elements meet the threshold. Final sum is 0.
Input
metrics = [10, 10, 10, 10], K = 10
Output
40
Explanation: Iterate through the array: 1. 10 >= 10: Add 10 to sum (sum = 10). 2. 10 >= 10: Add 10 to sum (sum = 20). 3. 10 >= 10: Add 10 to sum (sum = 30). 4. 10 >= 10: Add 10 to sum (sum = 40). All elements meet the threshold. Final sum is 40.
Input
metrics = [-5, -2, 0, 3, 7], K = -1
Output
10
Explanation: Iterate through the array: 1. -5 < -1: Skip. 2. -2 < -1: Skip. 3. 0 >= -1: Add 0 to sum (sum = 0). 4. 3 >= -1: Add 3 to sum (sum = 3). 5. 7 >= -1: Add 7 to sum (sum = 10). Final sum is 10.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum of all qualifying elements will fit within a 64-bit integer.
Optimal Approach & Strategy
The optimal approach is a single linear scan: iterate once, compare each metric to K, and add qualifying values to a running total, achieving O(n) time and O(1) space.
Brute Force Approach
A naive solution might sort the array first and then sum the tail, or use nested loops to filter and sum, leading to O(n log n) or O(n^2) time and extra memory usage.
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
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.