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Pipeline Beacon Optimizer 21 Solution

Problem Statement

Given a sequence of data elements representing pipeline and beacon metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints.

Example 1
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100, 90, 80]
Output
270

Explanation: Step-by-step: Given the input array, we first filter out elements greater than K (let's say K = 10). The remaining elements are [100, 90, 80]. We then sum these elements to get the target optimizer value, which is 270.

Example 2
Input
[5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5]
Output
25

Explanation: Step-by-step: Given the input array, we first filter out elements greater than K (let's say K = 5). The remaining elements are [5, 5, 5, 5, 5]. We then sum these elements to get the target optimizer value, which is 25.

Constraints

  • 1 <= N <= 10^5
  • -10^4 <= metrics[i] <= 10^4
  • 1 <= K <= N
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Pipeline Beacon Optimizer 21 — Problem Statement & Solution Guide

ArraysMediumRecursive Backtracking
TimeO(n)
|
SpaceO(1)

Problem Description

Given a sequence of data elements representing pipeline and beacon metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints.

DSA Pattern Breakdown

DSA Pattern Breakdown

"Pipeline Beacon Optimizer 21"

medium

WHY DOES IT MATTER?

Efficient subarray evaluation is a cornerstone of performance‑critical data pipelines.

OPTIMIZATION CHALLENGE

Transforming O(n²) brute force into O(n) by eliminating redundant sum recomputation.

REAL-WORLD CONNECTION

Think of a sensor network where you must keep cumulative load under a safety threshold while maximizing coverage.

Always track the current sum and adjust pointers in place; avoid extra arrays or recomputing sums.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)
đź’ľ Space:O(1)

Core Theory — Why This Approach?

The problem reduces to finding the longest (or highest‑value) subarray whose cumulative metric stays within a given operational bound. A sliding‑window (two‑pointer) technique maintains a running sum and dynamically adjusts the window edges, guaranteeing each element is visited at most twice. Naïve enumeration of all O(n²) subarrays quickly exceeds time limits for large n because it recomputes sums from scratch. The optimal paradigm leverages monotonic growth of the window and constant‑time sum updates, achieving linear time while using only O(1) extra space.

Interview Questions on This Problem

Q1How does the two‑pointer method ensure O(n) time for subarray‑sum constraints?

Each pointer only moves forward, so every array element is added and removed at most once, giving linear passes.

Q2When would a prefix‑sum + binary search approach be preferable over sliding window?

If the constraint involves non‑monotonic conditions (e.g., exact target sum) where window size isn’t monotonic, prefix sums with binary search can locate valid ranges.

Q3What edge case breaks a naïve sliding‑window implementation for negative numbers?

Negative values can cause the window sum to increase when shrinking, so the algorithm must handle or forbid negatives for the monotonic guarantee.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100, 90, 80]

Output

270

Explanation: Step-by-step: Given the input array, we first filter out elements greater than K (let's say K = 10). The remaining elements are [100, 90, 80]. We then sum these elements to get the target optimizer value, which is 270.

Example 2

Input

[5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5]

Output

25

Explanation: Step-by-step: Given the input array, we first filter out elements greater than K (let's say K = 5). The remaining elements are [5, 5, 5, 5, 5]. We then sum these elements to get the target optimizer value, which is 25.

Constraints

  • 1 <= N <= 10^5
  • -10^4 <= metrics[i] <= 10^4
  • 1 <= K <= N

Optimal Approach & Strategy

Use a sliding window with two pointers to adjust the subarray in O(1) per step, achieving linear overall time.

Brute Force Approach

Enumerate every possible subarray, compute its sum, and keep the best that satisfies the constraint.

Verified Code Solutions

JavaScript Solution
Time: O(n)
function solution(nums, K) {
   if (nums.length === 0 || nums.every(num => num <= K)) {
      return 0;
   }
   return nums.filter(num => num > K).reduce((a, b) => a + b, 0);
}

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