Pipeline Beacon Analyzer 24 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the throughput of a linear data pipeline. The pipeline consists of a sequence of processing nodes, each represented by an integer in an array beacons. Each integer denotes the signal strength emitted by that node. To ensure stable transmission, the system requires a contiguous segment of exactly k nodes to be analyzed simultaneously. Your objective is to identify the specific window of size k that yields the highest cumulative signal strength.
Given an array beacons of length n and an integer k, determine the maximum sum of any contiguous subarray of length k. If the array length is less than k, return -1, indicating that no valid window exists.
The solution must efficiently process the array to find the optimal window without resorting to brute-force recalculation for every position, leveraging the sliding window technique to maintain a constant time complexity per step after the initial window construction.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Analyzer 24"
WHY DOES IT MATTER?
Sliding‑window heap patterns turn per‑window O(k) work into logarithmic updates, crucial for real‑time analytics.
OPTIMIZATION CHALLENGE
The key is to avoid recomputing the maximum from scratch by maintaining a dynamic priority queue.
REAL-WORLD CONNECTION
Network routers often need the strongest signal in the last k packets to adjust transmission power.
Always store the element’s index alongside its value to know when it expires from the window.
COMPLEXITY AT A GLANCE
O(n log k)O(k)Core Theory — Why This Approach?
The sliding‑window maximum problem asks for the greatest element in every contiguous sub‑array of length k. A naïve scan recomputes the maximum for each window in O(k) time, leading to O(n·k) overall, which is prohibitive when n and k approach 10^5. The optimal paradigm leverages a max‑heap (or priority queue) that stores pairs (value, index). As the window slides, the heap’s top always holds the current maximum, while elements whose index falls outside the window are lazily removed, guaranteeing O(log k) updates per step. This approach reduces the total time to O(n log k) and uses O(k) auxiliary space, making it scalable for large streams.
Interview Questions on This Problem
Q1Why does a simple recomputation of the maximum for each window lead to TLE on large inputs?
It repeats O(k) work for each of the O(n) windows, resulting in O(n·k) operations which exceeds time limits when both are large.
Q2How does lazy deletion in a heap keep the structure size bounded to O(k)?
Elements are only physically removed when they reach the heap top; otherwise they stay but are ignored because their index is out of the current window.
Q3Can a deque replace the heap for this problem, and what are the trade‑offs?
Yes, a monotonic deque yields O(n) time with O(k) space, but a heap is more generic and easier to extend for variants like k‑th largest.
Examples
Input
beacons = [1, 2, 3, 4, 5], k = 3
Output
12
Explanation: The possible windows of size 3 are: [1, 2, 3] with sum 6, [2, 3, 4] with sum 9, and [3, 4, 5] with sum 12. The maximum sum is 12.
Input
beacons = [10, -5, 3, 8, 2], k = 4
Output
16
Explanation: The possible windows of size 4 are: [10, -5, 3, 8] with sum 16 and [-5, 3, 8, 2] with sum 8. The maximum sum is 16.
Input
beacons = [7, 7, 7, 7], k = 2
Output
14
Explanation: All windows of size 2 are [7, 7]. The sum for each is 14. The maximum sum is 14.
Input
beacons = [5], k = 2
Output
-1
Explanation: The array length is 1, which is less than k=2. No valid window exists, so the output is -1.
Constraints
- 1 <= beacons.length <= 10^5
- -10^9 <= beacons[i] <= 10^9
- 1 <= k <= 10^5
- The sum of all elements in any window will fit within a 64-bit integer.
Optimal Approach & Strategy
Maintain a max‑heap of the current window’s elements, inserting the new element and lazily removing out‑of‑range tops, achieving O(n log k) time.
Brute Force Approach
Iterate over each possible window and scan its k elements to find the maximum, costing O(n·k) time.
Verified Code Solutions
function maxSum(nums, k) {
if (k > nums.length) return 0;
let windowSum = 0;
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
windowSum += nums[i];
if (i >= k) windowSum -= nums[i - k];
if (i >= k - 1) maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}class Solution {
public:
int maxSum(vector<int>& nums, int k) {
if (k > nums.size()) return 0;
int windowSum = 0;
int maxSum = INT_MIN;
for (int i = 0; i < nums.size(); i++) {
windowSum += nums[i];
if (i >= k) windowSum -= nums[i - k];
if (i >= k - 1) maxSum = max(maxSum, windowSum);
}
return maxSum;
}
};public int maxSum(int[] nums, int k) {
if (k > nums.length) return 0;
int windowSum = 0;
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
windowSum += nums[i];
if (i >= k) windowSum -= nums[i - k];
if (i >= k - 1) maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}def max_sum(nums, k):
if k > len(nums):
return 0
window_sum = 0
max_sum = float('-inf')
for i in range(len(nums)):
window_sum += nums[i]
if i >= k:
window_sum -= nums[i - k]
if i >= k - 1:
max_sum = max(max_sum, window_sum)
return max_sumfunction maxSum(nums, k) {
if (k > nums.length) return 0;
let windowSum = 0;
let maxSum = -Infinity;
for (let i = 0; i < nums.length; i++) {
windowSum += nums[i];
if (i >= k) windowSum -= nums[i - k];
if (i >= k - 1) maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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