Pipeline Beacon Detector 35 — Problem Statement & Solution Guide

Dynamic ProgrammingMediumDFS Traversal

TimeO(n) for single query, O(n log n) preprocessing + O(log n) per query for multiple queries

SpaceO(1) for single query, O(n) for prefix sum array

Problem Description

In a distributed sensor network, a central controller receives a stream of integer readings from various pipeline nodes. Each reading represents a specific metric value. The system requires an analysis to identify the cumulative magnitude of all readings that exceed a critical safety threshold K. Your task is to implement a function that processes this sequence and returns the sum of all elements strictly greater than K. If no elements exceed the threshold, the function should return 0. This operation is fundamental for triggering high-priority alerts in real-time monitoring systems where only significant deviations matter.

DSA Pattern Breakdown

"Pipeline Beacon Detector 35"

medium

WHY DOES IT MATTER?

This pattern is essential because it represents the most common data processing task: filtering and aggregating. Understanding the trade-offs between linear scans, sorting + binary search, and advanced data structures (like Fenwick Trees) is critical for scaling systems from small scripts to distributed databases.

OPTIMIZATION CHALLENGE

The key insight is recognizing whether the threshold K is static or dynamic. If static, O(n) is optimal. If dynamic and multiple queries are expected, preprocessing with sorting and prefix sums reduces query time to O(log n). If updates are frequent, a Fenwick Tree offers O(log n) for both updates and queries.

REAL-WORLD CONNECTION

Think of a stock trading platform where you need to calculate the total volume of stocks trading above a certain price threshold. Or in IoT, summing the power consumption of all devices exceeding a safety limit. These are real-time aggregation tasks that require efficient computation.

In interviews, always clarify the constraints: Is the array static? Is K fixed? Are there multiple queries? This determines whether you propose a simple loop, a sorted array with binary search, or a complex data structure. Over-engineering a simple problem is a red flag, but under-engineering a complex one is worse.

COMPLEXITY AT A GLANCE

⏱ Time:O(n) for single query, O(n log n) preprocessing + O(log n) per query for multiple queries

💾 Space:O(1) for single query, O(n) for prefix sum array

Core Theory — Why This Approach?

The problem of summing elements that exceed a threshold K in a linear sequence is fundamentally a linear scan problem, often categorized under 'Filtering and Aggregation' patterns. While the prompt tags this as Dynamic Programming, the optimal solution for a static array is a single pass with O(n) time and O(1) space complexity. However, in the context of streaming data or dynamic updates (where elements are added or removed), this problem transitions into a Dynamic Programming or Data Structure problem. For instance, if the array is mutable and we need to query the sum of elements > K for arbitrary subarrays, we might use a Segment Tree or Fenwick Tree (Binary Indexed Tree), which are built on the principle of dynamic accumulation. The 'DP' aspect emerges when we consider the state of the sum as we traverse the array: let dp[i] be the sum of valid elements from index 0 to i. Then dp[i] = dp[i-1] + (arr[i] if arr[i] > K else 0). This recurrence relation is the hallmark of DP, even if it simplifies to a simple accumulator in the static case.

Interview Questions on This Problem

Q1At a fintech platform, we process millions of transaction records. How would you design a system to quickly calculate the total value of transactions exceeding a fraud detection threshold K, given that new transactions are added continuously?

For a continuous stream, a simple linear scan is insufficient for real-time queries. I would recommend a Fenwick Tree (BIT) or a Segment Tree if the values are bounded and we need range queries. If the threshold K is fixed and we only need the global sum, a running accumulator variable updated on each insert is O(1) per operation. If K changes dynamically, we might need a balanced BST (like a Treap) storing sums in subtrees to allow O(log n) updates and queries.

Q2In a distributed sensor network, nodes send readings in batches. How do you handle the case where the threshold K is not fixed but changes per query, and you need to answer multiple queries efficiently?

If K varies per query, a simple accumulator fails. I would sort the array once in O(n log n) time and compute a prefix sum array. For each query with threshold K, I can use binary search to find the first index where the value exceeds K, and then use the prefix sum to calculate the sum of the remaining elements in O(log n) time. This reduces the complexity from O(n) per query to O(log n) per query after O(n log n) preprocessing.

Q3You are building a high-growth startup's analytics dashboard. Users can filter data by 'value > K'. How do you optimize this for large datasets stored in a database?

In a database context, this is an index optimization problem. I would create an index on the value column. The query SELECT SUM(value) FROM table WHERE value > K would use the index to skip irrelevant rows. For in-memory processing, if the data is static, precomputing sorted arrays and prefix sums is best. If the data is dynamic, a B-Tree based structure (like in most RDBMS) allows efficient range scans and aggregations.

Examples

Example 1

Input

readings = [12, 5, 20, 8, 25], K = 10

Output

Explanation: Iterate through the readings: 12 > 10 (add 12), 5 <= 10 (skip), 20 > 10 (add 20), 8 <= 10 (skip), 25 > 10 (add 25). The sum is 12 + 20 + 25 = 57. Wait, let me re-calculate. 12+20+25 = 57. Let's adjust the example to be simpler or correct the math. Let's use [12, 5, 20, 8, 25] with K=10. 12>10, 20>10, 25>10. Sum = 12+20+25 = 57. Let's try another set to ensure variety. Let's use [3, 15, 7, 22, 4] with K=10. 15>10, 22>10. Sum = 37. Let's stick to the first one but correct the sum in the explanation. Actually, let's provide a clean example. Input: [12, 5, 20, 8, 25], K=10. Output: 57. Explanation: 12 is greater than 10, so add 12. 5 is not. 20 is greater than 10, so add 20. 8 is not. 25 is greater than 10, so add 25. Total sum = 12 + 20 + 25 = 57.

Example 2

Input

readings = [1, 2, 3, 4, 5], K = 10

Output

Explanation: Iterate through the readings: 1 <= 10, 2 <= 10, 3 <= 10, 4 <= 10, 5 <= 10. No elements are strictly greater than 10. Therefore, the sum remains 0.

Example 3

Input

readings = [100, 99, 98, 97, 96], K = 95

Output

Explanation: Iterate through the readings: 100 > 95 (add 100), 99 > 95 (add 99), 98 > 95 (add 98), 97 > 95 (add 97), 96 > 95 (add 96). The sum is 100 + 99 + 98 + 97 + 96 = 490.

Example 4

Input

readings = [-5, -10, 0, 5, 10], K = 0

Output

Explanation: Iterate through the readings: -5 <= 0, -10 <= 0, 0 <= 0 (not strictly greater), 5 > 0 (add 5), 10 > 0 (add 10). The sum is 5 + 10 = 15.

Constraints

1 <= readings.length <= 10^5
-10^9 <= readings[i] <= 10^9
-10^9 <= K <= 10^9

Optimal Approach & Strategy

For multiple queries with varying K, sort the array and compute a prefix sum array. For each query, use binary search to find the lower bound of K and use the prefix sum to calculate the sum of elements from that index to the end in O(log n) time.

Brute Force Approach

Iterate through the array and for each element, check if it is greater than K. If so, add it to a running total. This is actually the optimal solution for a single query on a static array, but if interpreted as a naive approach for multiple queries, it would be re-scanning the entire array for each query.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n) for single query, O(n log n) preprocessing + O(log n) per query for multiple queries

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); }

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Pipeline Beacon Detector 35 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Dynamic Programming

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Pipeline Beacon Detector 35 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor