Pipeline Beacon Aligner 27 — 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 aligner value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Aligner 27"
WHY DOES IT MATTER?
DP on DAG transforms an exponential path‑search into a linear scan.
OPTIMIZATION CHALLENGE
The key is reducing the combinatorial explosion of paths to a single pass over edges.
REAL-WORLD CONNECTION
Similar to scheduling tasks with dependencies where the earliest finish time is computed.
Always verify acyclicity first; a hidden cycle silently breaks the DP logic.
COMPLEXITY AT A GLANCE
O(V+E)O(V)Core Theory — Why This Approach?
The problem can be modeled as a directed graph where vertices represent pipeline segments or beacons and edges encode permissible alignment transitions. Because the operational constraints enforce a strict ordering, the graph is a Directed Acyclic Graph (DAG), allowing us to compute the optimal aligner value with a single pass of dynamic programming over a topological ordering. A naive exhaustive search would enumerate every possible path, leading to exponential time (O(2^N)) and quickly exhausting memory on large inputs. By exploiting the DAG property, we replace exponential recursion with linear DP: each vertex stores the best aligner value achievable up to that point, and we propagate these values along edges in topological order, guaranteeing optimality in O(V+E) time.
Interview Questions on This Problem
Q1How do you detect whether the given graph is a DAG and why is this check important for the solution?
Perform a topological sort using Kahn's algorithm or DFS cycle detection; if the sort fails, a cycle exists and the DP approach is invalid.
Q2Explain how dynamic programming on a topologically sorted DAG yields the optimal aligner value.
Processing vertices in topological order ensures all predecessors of a node have already computed their best values, so we can update the node by taking the maximum of predecessor values plus its own metric.
Q3What is the time and space complexity of the DP‑on‑DAG solution and how does it compare to the brute‑force approach?
The DP solution runs in O(V+E) time and O(V) space, whereas brute force explores all paths in exponential time and uses exponential extra memory.
Examples
Input
[1, 2, 3, 4, 5] and K = 3
Output
0
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and K = 3, we iterate through the array. Since all elements are less than or equal to K, we return 0 as the sum of elements greater than K is 0.
Input
[10, 20, 30, 40, 50] and K = 25
Output
0
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50] and K = 25, we iterate through the array. Since all elements are less than or equal to K, we return 0 as the sum of elements greater than K is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Topologically sort the DAG and apply DP to compute the best value for each vertex in a single linear pass.
Brute Force Approach
Enumerate every possible path from start to end and compute its aligner value, which is exponential in the number of vertices.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > k) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
}
}
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 i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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