Pipeline Grid Extractor 2 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and grid metrics, and an integer K, construct an optimal algorithm to evaluate and compute the target extractor value under the given operational constraints. The target extractor value is calculated as the sum of the subarray of size K with the maximum sum, squared, plus K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Grid Extractor 2"
WHY DOES IT MATTER?
The sliding‑window (queue) pattern turns a potentially quadratic scan into a linear pass, which is essential for real‑time analytics and high‑throughput systems where latency must stay sub‑millisecond per element.
OPTIMIZATION CHALLENGE
The key insight is recognizing that adjacent windows differ by exactly two elements—one entering, one leaving—so the window sum can be updated by subtracting the outgoing element and adding the incoming one, eliminating the need for recomputation.
REAL-WORLD CONNECTION
Think of a moving average filter on a sensor stream: as new readings arrive, the oldest reading exits the window. Maintaining the sum in a queue mirrors how distributed monitoring services compute rolling metrics without re‑processing the entire history.
During an interview, write the sliding‑window loop first, then immediately add a variable to track the maximum sum; this keeps the code concise and avoids a second pass over the data.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem asks for the maximum‑sum subarray of a fixed length K, a classic sliding‑window scenario. A naïve solution would recompute the sum for every possible window, leading to O(N·K) time, which quickly becomes prohibitive when N (the number of pipeline metrics) reaches 10^5 or more. The optimal paradigm leverages the fact that consecutive windows overlap by K‑1 elements; by maintaining a running sum and updating it in O(1) as the window slides, we achieve linear time. Once the maximum sum is identified, the final extractor value is simply (maxSum)^2 + K, a constant‑time arithmetic step.
The sliding‑window technique is a special case of a queue‑based approach: the current window can be thought of as a queue where we enqueue the next element and dequeue the element that falls out of the window. This abstraction clarifies why only O(1) work per step is needed and why the algorithm naturally runs in O(N) time and O(1) auxiliary space. Understanding this pattern also prepares candidates for many related problems such as longest subarray with constraints, minimum window substring, and real‑time stream analytics.
Interview Questions on This Problem
Q1How would you modify the solution if the window size K could be any value up to N and you needed to answer multiple queries for different K efficiently?
Pre‑compute prefix sums for the array; then each query for a specific K can be answered in O(N) by scanning the prefix‑sum differences, or in O(1) per window using a deque to maintain maximum sums for all K simultaneously, which leads to an O(N) preprocessing and O(1) per query solution.
Q2Why is a sliding‑window preferable to a priority queue for this problem?
A priority queue would require O(log K) insertion and removal for each step, inflating the total to O(N log K). The sliding‑window only needs constant‑time updates because the window’s composition changes predictably, giving a strict O(N) bound.
Q3Explain how you would handle negative numbers in the array and still guarantee the correct maximum‑sum window.
The sliding‑window algorithm works unchanged with negative values because it always tracks the exact sum of the current K‑length segment; the maximum sum may be negative, but the algorithm still records the highest (least negative) sum encountered.
Examples
Input
[40, 50], 2
Output
8102
Explanation: Step-by-step: with input [40, 50] and K = 2, we find the subarray of size K with the maximum sum, which is [40, 50] with a sum of 90. Then, we calculate the target extractor value as 90 * 90 + 2 = 8100 + 2 = 8102.
Input
[30, 40, 50], 3
Output
14403
Explanation: Step-by-step: with input [30, 40, 50] and K = 3, we find the subarray of size K with the maximum sum, which is [30, 40, 50] with a sum of 120. Then, we calculate the target extractor value as 120 * 120 + 3 = 14400 + 3 = 14403.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a sliding window (queue) to maintain the current K‑element sum, updating it in O(1) as the window moves, while tracking the maximum sum encountered.
Brute Force Approach
Compute the sum of every possible subarray of length K by iterating over all start indices and summing K elements each time.
Verified Code Solutions
function solution(nums, k) {
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - k; i++) {
let sum = 0;
for (let j = i; j < i + k; j++) {
sum += nums[j];
}
maxSum = Math.max(maxSum, sum);
}
return maxSum * maxSum + k;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int maxSum = INT_MIN;
for (int i = 0; i <= nums.size() - k; i++) {
int sum = 0;
for (int j = i; j < i + k; j++) {
sum += nums[j];
}
maxSum = max(maxSum, sum);
}
return maxSum * maxSum + k;
}
};class Solution {
public int solution(int[] nums, int k) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i <= nums.length - k; i++) {
int sum = 0;
for (int j = i; j < i + k; j++) {
sum += nums[j];
}
maxSum = Math.max(maxSum, sum);
}
return maxSum * maxSum + k;
}
}def solution(nums, k):
max_sum = float('-inf')
for i in range(len(nums) - k + 1):
subarray_sum = sum(nums[i:i+k])
max_sum = max(max_sum, subarray_sum)
return max_sum * max_sum + kfunction solution(nums, k) {
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - k; i++) {
let sum = 0;
for (let j = i; j < i + k; j++) {
sum += nums[j];
}
maxSum = Math.max(maxSum, sum);
}
return maxSum * maxSum + k;
}Asked in Top Tech Interviews
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