Pipeline Grid Detector 23 — Problem Statement & Solution Guide

Two PointersMediumFrequency Hash Map

TimeO(n)

SpaceO(1)

Problem Description

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

DSA Pattern Breakdown

"Pipeline Grid Detector 23"

medium

WHY DOES IT MATTER?

Two‑pointer patterns turn quadratic subarray problems into linear solutions, crucial for scaling to millions of elements.

OPTIMIZATION CHALLENGE

The main challenge is maintaining the invariant while moving pointers without recomputing the whole window.

REAL-WORLD CONNECTION

It mirrors how network routers slide a congestion window to adapt to bandwidth constraints.

Always track the aggregate metric incrementally; avoid recalculating sums from scratch inside the loop.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The two‑pointer (or sliding‑window) technique transforms a problem that appears to require O(n²) pairwise checks into a linear scan by maintaining a dynamic interval whose endpoints move monotonically. By expanding the right pointer to include new elements and contracting the left pointer only when the current window violates the constraint, we guarantee each element is visited at most twice, achieving O(n) time.

Naïve brute‑force enumerates every possible start‑end pair, recomputing sums or metrics from scratch, which explodes to O(n²) or worse for large n and exceeds time limits. The optimal paradigm leverages the monotonic nature of the constraint (e.g., non‑decreasing sum, bounded difference) so that once a window is invalid, moving the left pointer forward restores validity without revisiting earlier elements, ensuring both correctness and efficiency.

Interview Questions on This Problem

Q1How does the sliding‑window guarantee O(n) time for problems with a sum constraint?

Each element is added to the window once when the right pointer moves and removed once when the left pointer moves. This bounded movement means the total number of pointer operations is at most 2n.

Q2When would a two‑pointer approach fail for a given problem?

If the constraint is not monotonic—e.g., requires checking arbitrary subsets—the window cannot be adjusted deterministically. In such cases, more complex data structures or DP may be needed.

Q3What is the key difference between a fixed‑size window and a variable‑size window?

A fixed‑size window slides with a constant length, useful for exact‑length subarrays. A variable‑size window expands or contracts based on a condition, enabling optimization of length or sum.

Examples

Example 1

Input

[10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

Output

Explanation: Step-by-step: 1. Sort the array in descending order. 2. Select the K largest values from the sorted array. 3. Sum the selected values. 4. If K is larger than the number of unique values in the array, add the next largest value to the sum. For the given input, the K largest values are 10, 9, 8, 7, 6. The sum is 10 + 9 + 8 + 7 + 6 = 40.

Example 2

Input

[10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

Output

Constraints

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

Optimal Approach & Strategy

Use a sliding window with left and right pointers, updating the metric incrementally and adjusting the window only when the constraint is violated.

Brute Force Approach

Check every possible start and end index, recompute the metric for each subarray, and keep the best result.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, k) {
   nums.sort((a, b) => b - a);
   let sum = 0;
   for (let i = 0; i < k; i++) {
       sum += nums[i];
   }
   if (k > nums.length) {
       sum += nums[k - 1];
   }
   return sum;
}

class Solution {
   public:
       int solution(vector<int>& nums, int k) {
           sort(nums.begin(), nums.end(), greater<int>());
           int sum = 0;
           for (int i = 0; i < k; i++) {
               sum += nums[i];
           }
           if (k > nums.size()) {
               sum += nums[k - 1];
           }
           return sum;
       }
   };

class Solution {
   public int solution(int[] nums, int k) {
       Arrays.sort(nums);
       int sum = 0;
       for (int i = 0; i < k; i++) {
           sum += nums[i];
       }
       if (k > nums.length) {
           sum += nums[k - 1];
       }
       return sum;
   }
}

def solution(nums, k):
   nums.sort(reverse=True)
   sum = 0
   for i in range(k):
       sum += nums[i]
   if k > len(nums):
       sum += nums[k - 1]
   return sum

function solution(nums, k) {
   nums.sort((a, b) => b - a);
   let sum = 0;
   for (let i = 0; i < k; i++) {
       sum += nums[i];
   }
   if (k > nums.length) {
       sum += nums[k - 1];
   }
   return sum;
}

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Pipeline Grid Detector 23 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Pipeline Grid Detector 23 — 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