Protocol Sensor Extractor 43 — 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 extractor value under given operational constraints. The operational constraints are that the maximum value in the array should be less than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Extractor 43"
WHY DOES IT MATTER?
Monotonic queues turn a potentially quadratic max‑query into linear time, a staple for real‑time stream processing.
OPTIMIZATION CHALLENGE
The key is reducing repeated max scans to a single pass by amortizing deque operations.
REAL-WORLD CONNECTION
Network routers use similar structures to keep track of peak traffic rates over sliding time windows.
Always purge stale indices before reading the deque front; forgetting this creates subtle bugs.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The optimal solution relies on the sliding‑window technique combined with a monotonic decreasing deque to track the current window's maximum in O(1) amortized time. A naive double‑loop that checks every sub‑array would be O(N²) and fails for N up to 10⁵, whereas the deque ensures each element is inserted and removed at most once, yielding linear complexity.
Interview Questions on This Problem
Q1How does a monotonic deque help maintain the maximum of a sliding window?
It stores indices in decreasing order of their values, so the front always holds the current maximum. Out‑of‑range indices are popped from the front, and smaller values are removed from the back on insertion.
Q2Why is the two‑pointer (left‑right) approach essential for this problem?
It lets us expand the window until the constraint (max ≤ K) is violated, then contract from the left to restore validity, guaranteeing each element is processed a constant number of times.
Q3What is the overall time and space complexity of the deque‑based solution?
Both time and space are O(N) in the worst case because each array element is pushed and popped at most once from the deque.
Examples
Input
[120, 100, 80, 95, 70]
Output
120
Explanation: Step-by-step: Given the input [120, 100, 80, 95, 70], we first find the maximum value in the array, which is 120. Since 120 is greater than K=100, we return the maximum value in the array, which is 120.
Input
[95, 80, 70, 60, 50]
Output
95
Explanation: Step-by-step: Given the input [95, 80, 70, 60, 50], we first find the maximum value in the array, which is 95. Since 95 is less than or equal to K=80, we return the maximum value in the array, which is 95.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain left/right pointers and a monotonic deque to get the current max in O(1), adjusting the left pointer when max>K – O(N) time.
Brute Force Approach
Check every possible sub‑array, compute its max, and keep the longest valid one – O(N²) time.
Verified Code Solutions
function solution(nums, K) {
let max = Math.max(...nums);
return max <= K ? max : Math.max(...nums);
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int max = INT_MIN;
for (int num : nums) {
max = max > num ? max : num;
}
return max <= K ? max : max;
}
};class Solution {
public int solution(int[] nums, int K) {
int max = Integer.MAX_VALUE;
for (int num : nums) {
max = Math.max(max, num);
}
return max <= K ? max : max;
}
}def solution(nums, K):
max_val = max(nums)
return max_val <= K and max_val or max(nums)function solution(nums, K) {
let max = Math.max(...nums);
return max <= K ? max : Math.max(...nums);
}Asked in Top Tech Interviews
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