Pipeline Beacon Architect 11 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and beacon metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints. The input array nums represents the pipeline and beacon metrics, and the integer k represents the operational constraint. The solution should return the sum of elements in nums that are greater than k.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Architect 11"
WHY DOES IT MATTER?
Monotonic queues turn a seemingly quadratic problem into linear time, a core optimization pattern.
OPTIMIZATION CHALLENGE
The key is reducing redundant comparisons by discarding dominated elements as the window moves.
REAL-WORLD CONNECTION
Network routers use similar sliding‑window techniques to track peak traffic over recent intervals.
Always clean the deque's front before recording the result to avoid stale indices.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The sliding‑window maximum problem asks for the greatest element in every contiguous subarray of length k. A naïve double loop examines each window independently, yielding O(n·k) time, which explodes for large n (e.g., n = 10⁶) and is unacceptable in interview settings. The optimal paradigm uses a double‑ended queue (deque) to store indices of candidates in decreasing order; as the window slides, elements that fall out are popped from the front and smaller elements are discarded from the back, guaranteeing each element is inserted and removed at most once, thus achieving linear time. This approach leverages the monotonic queue invariant to maintain the current maximum in O(1) per step while preserving O(n) overall complexity.
Interview Questions on This Problem
Q1Why does a deque enable O(n) time for sliding‑window maximum?
Because each array element is pushed and popped at most once, maintaining a monotonic decreasing order of candidates.
Q2How would you adapt the sliding‑window maximum algorithm to a singly linked list?
Traverse the list while storing node references in a deque, discarding nodes that exit the window and maintaining the monotonic property.
Q3What edge case must you handle when k equals 1 or n?
When k = 1, every element is its own maximum; when k = n, the answer is simply the global maximum of the array.
Examples
Input
[5, 6, 7, 8, 9, 10]
Output
40
Explanation: Step-by-step: Given the input array [5, 6, 7, 8, 9, 10] and k = 5, we iterate through the array and sum up all elements greater than k. In this case, the elements 6, 7, 8, 9, and 10 are greater than 5, so the sum is 6 + 7 + 8 + 9 + 10 = 40.
Input
[60, 70, 80, 90, 100, 100]
Output
340
Explanation: Step-by-step: Given the input array [60, 70, 80, 90, 100, 100] and k = 60, we iterate through the array and sum up all elements greater than k. In this case, the elements 70, 80, 90, 100, and 100 are greater than 60, so the sum is 70 + 80 + 90 + 100 + 100 = 340.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a decreasing deque of indices, updating it as the window slides to achieve O(n) time.
Brute Force Approach
Iterate over each window and scan k elements to find the max, resulting in O(n·k) time.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int num : nums) {
if (num > k) {
sum += num;
}
}
return sum;
}
}def solution(nums, k):
sum = 0
for num in nums:
if num > k:
sum += num
return sumfunction solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num > k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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