Pipeline Beacon Evaluator 38 — Problem Statement & Solution Guide

Two PointersMediumGreedy Choice

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 evaluator value under given operational constraints.

DSA Pattern Breakdown

"Pipeline Beacon Evaluator 38"

medium

WHY DOES IT MATTER?

The two-pointer pattern reduces time complexity from quadratic to linear for problems involving contiguous segments, making large datasets tractable. It also simplifies reasoning about the state of the algorithm, as the window boundaries directly reflect the current subproblem.

OPTIMIZATION CHALLENGE

The core insight is that the sum of the window can be updated in O(1) when moving either pointer, eliminating the need to recompute sums from scratch. This transforms an O(n²) brute force into O(n).

REAL-WORLD CONNECTION

In distributed systems, a sliding window is used for rate limiting: the window represents a time interval, and the system tracks the number of requests within that interval to enforce limits. Adjusting the window as time progresses mirrors the pointer adjustments in the algorithm.

When explaining the algorithm, emphasize the invariant that the window always satisfies the sum constraint after the left pointer moves. This helps interviewers see that the algorithm is correct and efficient.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

Two pointers, often called the sliding window technique, is a paradigm that maintains two indices that traverse an array or string to maintain a contiguous segment that satisfies a particular property. In the context of finding the longest subarray with sum ≤ k, a naive approach would examine every possible start and end pair, leading to an O(n²) time complexity. By maintaining a window [left, right] and expanding the right pointer while the sum stays within the limit, we can adjust the left pointer to shrink the window when the sum exceeds k. This ensures each element is added and removed at most once, yielding an O(n) solution. The key insight is that the property (sum ≤ k) is monotonic with respect to window expansion: adding an element can only increase the sum, and removing an element can only decrease it, allowing us to adjust pointers deterministically.

Interview Questions on This Problem

Q1How would you modify the sliding window algorithm if the array contains negative numbers?

With negative numbers, the sum can decrease when expanding the window, so the monotonic property no longer holds. One approach is to use a prefix sum array and a balanced BST or hash map to track the earliest index where the prefix sum is at least currentSum - k, achieving O(n log n) or O(n) with a deque for specific constraints.

Q2Can you explain a real-world scenario where a two-pointer technique would be useful outside of arrays?

In network packet processing, you might need to find the longest contiguous sequence of packets whose total size does not exceed a bandwidth limit. A sliding window over the packet sizes can efficiently maintain the current total and adjust the window as new packets arrive.

Q3What are the pitfalls when implementing the sliding window for the longest subarray sum problem, and how would you test for them?

Common pitfalls include off-by-one errors when moving pointers, failing to reset the sum when the window becomes empty, and not handling the case where all elements exceed k. Unit tests should cover arrays with all positive numbers, arrays with a single element > k, empty arrays, and arrays where the optimal subarray is at the beginning or end.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 3

Output

Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 3, we use two pointers to find the first pair of elements greater than K. The first pointer starts at the beginning of the array, and the second pointer starts at the end of the array. We move the second pointer to the left until we find a pair of elements greater than K. The output is the sum of these two elements, which is 4 + 5 = 9. However, we are asked to find the count of such pairs, not their sum. Therefore, the correct output is 2, not 9. However, the problem asks for the count of such pairs, not their sum. Therefore, the correct output is 2, not 9.

Example 2

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 25

Output

Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 25, we use two pointers to find the first pair of elements greater than K. However, since all elements in the array are less than or equal to K, there are no pairs of elements greater than K. Therefore, the output is 0.

Constraints

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

Optimal Approach & Strategy

Use a sliding window with two pointers: expand the right pointer while the sum stays ≤ k, and when it exceeds k, move the left pointer to reduce the sum. Track the maximum window length seen. This runs in O(n) time and O(1) space.

Brute Force Approach

Check every possible subarray by nested loops, compute its sum, and update the maximum length if the sum is ≤ k. This takes O(n²) time and O(1) space.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, K) {
      let count = 0;
      let left = 0;
      let right = nums.length - 1;
      while (left < right) {
         if (nums[left] + nums[right] > K) {
            count++;
            left++;
         } else {
            right--;
         }
      }
      return count;
   }

class Solution {
   public:
      int solution(vector<int>& nums, int K) {
         int count = 0;
         int left = 0;
         int right = nums.size() - 1;
         while (left < right) {
            if (nums[left] + nums[right] > K) {
               count++;
               left++;
            } else {
               right--;
            }
         }
         return count;
      }
   };

class Solution {
      public int solution(int[] nums, int K) {
         int count = 0;
         int left = 0;
         int right = nums.length - 1;
         while (left < right) {
            if (nums[left] + nums[right] > K) {
               count++;
               left++;
            } else {
               right--;
            }
         }
         return count;
      }
   }

def solution(nums, K):
      count = 0
      left = 0
      right = len(nums) - 1
      while left < right:
         if nums[left] + nums[right] > K:
            count += 1
            left += 1
         else:
            right -= 1
      return count

function solution(nums, K) {
      let count = 0;
      let left = 0;
      let right = nums.length - 1;
      while (left < right) {
         if (nums[left] + nums[right] > K) {
            count++;
            left++;
         } else {
            right--;
         }
      }
      return count;
   }

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Pipeline Beacon Evaluator 38 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Pipeline Beacon Evaluator 38 — 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