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Protocol Pipeline Tracker 15 Solution

Problem Statement

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

Example 1
Input
[6, 7, 8, 9, 10], 5
Output
5

Explanation: Step-by-step: Given the input array [6, 7, 8, 9, 10] and the target K = 5, we iterate over the array and add up all numbers greater than K. In this case, we add 6, 7, 8, 9, and 10, giving us a total sum of 5.

Example 2
Input
[1, 2, 3, 4, 5], 5
Output
0

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and the target K = 5, we iterate over the array and add up all numbers greater than K. In this case, we do not add any numbers since all numbers are less than or equal to K, giving us a total sum of 0.

Constraints

  • 1 <= N <= 10^5
  • -10^4 <= metrics[i] <= 10^4
  • 1 <= K <= N
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Protocol Pipeline Tracker 15 — Problem Statement & Solution Guide

Sliding WindowEasyBitmasking
TimeO(n)
|
SpaceO(1)

Problem Description

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

DSA Pattern Breakdown

DSA Pattern Breakdown

"Protocol Pipeline Tracker 15"

easy

WHY DOES IT MATTER?

Sliding windows turn quadratic subarray scans into linear passes, essential for real‑time analytics.

OPTIMIZATION CHALLENGE

The key is reducing repeated work by updating aggregates incrementally instead of recomputing them.

REAL-WORLD CONNECTION

Network routers use sliding windows to track packet loss rates over the last N seconds.

Keep the window state minimal—store only what you need (sum, count, deque) to avoid hidden O(k) overhead.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)
đź’ľ Space:O(1)

Core Theory — Why This Approach?

The sliding window technique transforms a problem that naïvely requires O(n·k) time—by recomputing aggregates for every possible sub‑segment—into a linear scan by maintaining a dynamic window of relevant elements. By updating the window’s state (sum, max, count, etc.) as the left and right pointers move, each element is processed a constant number of times, guaranteeing O(n) overall complexity. Naïve approaches fail on large inputs because they repeatedly traverse overlapping portions of the array, leading to quadratic time and memory blow‑up. The optimal paradigm leverages the monotonicity or additive properties of the metric, allowing incremental updates and immediate discarding of stale data, which is the essence of the sliding window pattern.

Interview Questions on This Problem

Q1How does the sliding window achieve O(n) time for sum‑based window problems?

It adds the incoming element and subtracts the outgoing element as the window slides, so each element is touched twice at most. This eliminates the need to recompute the sum from scratch for each position.

Q2When would you prefer a deque over a simple two‑pointer approach?

A deque efficiently maintains monotonic information like max/min within the window. It allows O(1) access to the extreme value while still supporting O(1) insertions and deletions.

Q3What edge case can break a sliding‑window implementation that assumes a fixed window size?

Variable‑size windows (e.g., “minimum length subarray with sum ≥ S”) require dynamic adjustment of the left pointer. Forgetting to shrink the window when the condition is met leads to incorrect results or infinite loops.

Examples

Example 1

Input

[6, 7, 8, 9, 10], 5

Output

5

Explanation: Step-by-step: Given the input array [6, 7, 8, 9, 10] and the target K = 5, we iterate over the array and add up all numbers greater than K. In this case, we add 6, 7, 8, 9, and 10, giving us a total sum of 5.

Example 2

Input

[1, 2, 3, 4, 5], 5

Output

0

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and the target K = 5, we iterate over the array and add up all numbers greater than K. In this case, we do not add any numbers since all numbers are less than or equal to K, giving us a total sum of 0.

Constraints

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

Optimal Approach & Strategy

Use two pointers to expand and shrink a window while updating the metric incrementally, achieving linear time.

Brute Force Approach

Iterate over every possible subarray, recompute the metric from scratch, and track the best result.

Verified Code Solutions

JavaScript Solution
Time: O(n)
function solution(nums, K) {
   let sum = 0;
   for (let num of nums) {
       if (num > K) {
           sum += num;
       }
   }
   return sum;
}

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