Pipeline Vector Synthesizer 41 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and vector metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Vector Synthesizer 41"
WHY DOES IT MATTER?
Efficient path aggregation on weighted graphs is a core pattern for many optimization problems.
OPTIMIZATION CHALLENGE
The key is reducing exponential path enumeration to linear‑time DP or greedy edge relaxations.
REAL-WORLD CONNECTION
It mirrors routing decisions in data‑center pipelines where latency or bandwidth metrics must be aggregated.
Always verify DAG properties first; a quick cycle check can save you from picking the wrong algorithm.
COMPLEXITY AT A GLANCE
O(N + M)O(N + M)Core Theory — Why This Approach?
The problem can be abstracted as a directed graph where each data element is a vertex and the pipeline‑vector relationships are directed edges with weights representing metric contributions. Computing the target synthesizer value is equivalent to finding the maximum (or minimum) accumulated weight along any feasible path that respects the operational constraints, which often translate to a DAG or a graph with non‑negative edge weights. A naive exhaustive search that enumerates all paths quickly becomes exponential (O(2^N) in the worst case) and fails for large N because the number of possible traversals grows combinatorially. The optimal paradigm leverages graph‑theoretic properties: if the graph is acyclic, a topological sort yields a linear ordering that allows a single‑pass dynamic programming sweep, achieving O(N+M) time; if cycles exist but edge weights are non‑negative, Dijkstra’s algorithm with a priority queue provides the same asymptotic bound while handling arbitrary directed graphs. Both approaches avoid redundant recomputation by storing the best known value for each vertex and only relaxing edges when an improvement is possible.
Interview Questions on This Problem
Q1How do you decide whether to use topological DP versus Dijkstra for this problem?
If the graph is guaranteed to be a DAG, topological DP is simpler and runs in linear time. Otherwise, when cycles may exist but weights are non‑negative, Dijkstra’s algorithm is the safe choice.
Q2What is the purpose of edge relaxation in these algorithms?
Relaxation updates the best known value for a destination vertex using a candidate path through an edge. It ensures that after processing all edges, each vertex holds the optimal accumulated metric.
Q3Why is it important to detect disconnected components before processing?
Disconnected vertices cannot contribute to the target value and may cause incorrect initialization if left unchecked. Handling them early lets the algorithm return a default or ignore them safely.
Examples
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 20, 30]]
Output
6
Explanation: Step 1: Initialize the sum to 0. Step 2: Iterate over each sub-array in the input. Step 3: For each sub-array, find the maximum value and add it to the sum. Step 4: Return the final sum.
Input
[[1, 2, 3], [4, 5, 6], [7, 8, 9], []]
Output
0
Explanation: Step 1: Initialize the sum to 0. Step 2: Iterate over each sub-array in the input. Step 3: If the sub-array is not empty, find the maximum value and add it to the sum. Step 4: Return the final sum.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Perform a topological sort (or Dijkstra) and apply dynamic programming to propagate the best metric value along edges in O(N+M) time.
Brute Force Approach
Enumerate every possible path and sum its metrics, which is exponential and infeasible for large inputs.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i].length > 0) {
let max = Math.max(...nums[i]);
sum += max;
}
}
return sum;
}class Solution {
public:
int solution(vector<vector<int>>& nums) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (!nums[i].empty()) {
int max = INT_MIN;
for (int j = 0; j < nums[i].size(); j++) {
max = max(max, nums[i][j]);
}
sum += max;
}
}
return sum;
}
};class Solution {
public int solution(int[][] nums) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i].length > 0) {
int max = Integer.MAX_VALUE;
for (int j = 0; j < nums[i].length; j++) {
max = Math.max(max, nums[i][j]);
}
sum += max;
}
}
return sum;
}
}def solution(nums):
sum = 0
for i in range(len(nums)):
if nums[i]:
max_val = max(nums[i])
sum += max_val
return sumfunction solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i].length > 0) {
let max = Math.max(...nums[i]);
sum += max;
}
}
return sum;
}Asked in Top Tech Interviews
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