Protocol Sensor Detector 30 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and sensor metrics, construct an optimal algorithm to evaluate and compute the target detector value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Detector 30"
WHY DOES IT MATTER?
Sliding windows turn quadratic subarray scans into linear passes, crucial for real‑time data streams.
OPTIMIZATION CHALLENGE
The key is maintaining the window’s aggregate in O(1) while adjusting boundaries, cutting the complexity from O(n²) to O(n).
REAL-WORLD CONNECTION
Network routers use a moving window to monitor packet rates and trigger alerts when thresholds are crossed.
Initialize pointers outside the loop, update aggregates before moving pointers, and guard against empty‑window edge cases.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The sliding window technique transforms a problem that naively requires examining all sub‑arrays—resulting in O(n²) time—into a linear scan by maintaining a dynamic range that satisfies the problem’s constraints. By incrementally adding the next element to the window and, when necessary, shrinking from the left, we preserve the invariant (e.g., sum ≤ limit) while visiting each element at most twice.
In the context of the Protocol Sensor Detector, the window represents the current contiguous segment of protocol and sensor metrics under evaluation. Updating aggregate metrics (such as sum, max, or count) in O(1) per step eliminates redundant recomputation, delivering an optimal O(n) time algorithm with O(1) auxiliary space, which scales to massive input streams where brute‑force would time out.
Interview Questions on This Problem
Q1How does the sliding window technique achieve O(n) time for contiguous subarray problems?
Each element enters and exits the window at most once, so the total number of operations is bounded by 2n. This eliminates the nested loops of the naive approach.
Q2When would you prefer a variable‑size window over a fixed‑size window?
A variable‑size window is needed when the constraint depends on the window’s content (e.g., sum ≤ K) rather than its length. Fixed‑size windows are optimal for problems with a predetermined length.
Q3What pitfalls arise when updating aggregates like the sum while shrinking the window?
Failing to subtract the leaving element or handling integer overflow can corrupt the invariant. Always update the aggregate before moving the left pointer.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
45
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we calculate the maximum sum of all elements within the window of size 3. The window is [1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 6], [5, 6, 7], [6, 7, 8], [7, 8, 9], [8, 9, 10]. The maximum sum is 15 + 16 + 18 + 20 + 22 + 24 + 26 + 28 = 159. However, we need to consider the maximum sum of all subarrays of size 3. The maximum sum is 15 + 16 + 18 = 49, 16 + 18 + 20 = 54, 18 + 20 + 22 = 60, 20 + 22 + 24 = 66, 22 + 24 + 26 = 72, 24 + 26 + 28 = 78, 26 + 28 + 30 = 84, 28 + 30 + 32 = 90. The maximum sum is 90.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
Output
120
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], we calculate the maximum sum of all elements within the window of size 5. The window is [1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [3, 4, 5, 6, 7], [4, 5, 6, 7, 8], [5, 6, 7, 8, 9], [6, 7, 8, 9, 10], [7, 8, 9, 10, 11], [8, 9, 10, 11, 12], [9, 10, 11, 12, 13], [10, 11, 12, 13, 14], [11, 12, 13, 14, 15]. The maximum sum is 15 + 16 + 17 + 18 + 19 = 85, 16 + 17 + 18 + 19 + 20 = 90, 17 + 18 + 19 + 20 + 21 = 95, 18 + 19 + 20 + 21 + 22 = 100, 19 + 20 + 21 + 22 + 23 = 105, 20 + 21 + 22 + 23 + 24 = 110, 21 + 22 + 23 + 24 + 25 = 115, 22 + 23 + 24 + 25 + 26 = 120, 23 + 24 + 25 + 26 + 27 = 125, 24 + 25 + 26 + 27 + 28 = 130, 25 + 26 + 27 + 28 + 29 = 135, 26 + 27 + 28 + 29 + 30 = 140. The maximum sum is 140.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain two pointers defining the current window, adjust the right pointer to include new elements, and move the left pointer when the constraint is violated, achieving O(n) time.
Brute Force Approach
Iterate over every possible start index and, for each, expand to all end indices while recomputing the metric, leading to O(n²) time.
Verified Code Solutions
function solution(nums, k) {
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - k; i++) {
let windowSum = 0;
for (let j = i; j < i + k; j++) {
windowSum += nums[j];
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int max_sum = INT_MIN;
for (int i = 0; i <= nums.size() - k; i++) {
int window_sum = 0;
for (int j = i; j < i + k; j++) {
window_sum += nums[j];
}
max_sum = max(max_sum, window_sum);
}
return max_sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i <= nums.length - k; i++) {
int windowSum = 0;
for (int j = i; j < i + k; j++) {
windowSum += nums[j];
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}
}def solution(nums, k):
max_sum = float('-inf')
for i in range(len(nums) - k + 1):
window_sum = sum(nums[i:i + k])
max_sum = max(max_sum, window_sum)
return max_sumfunction solution(nums, k) {
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - k; i++) {
let windowSum = 0;
for (let j = i; j < i + k; j++) {
windowSum += nums[j];
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}Asked in Top Tech Interviews
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