How to Solve Network Recovery Pathways Leetcode Problem - Interview Coder Guide
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How to Solve Network Recovery Pathways Problem

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Hard#3919
LeetCode Problem

Network Recovery Pathways

You are given a directed acyclic graph of n nodes numbered from 0 to n − 1. This is represented by a 2D array edges of length m, where edges[i] = [ui, vi, costi] indicates a one‑way communication from...

ArrayBinary SearchDynamic Programming+4 more

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Problem Breakdown

Understanding the Network Recovery Pathways Problem

Let's break down this LeetCode problem and understand what makes it challenging in interview settings.

Problem Statement

You are given a directed acyclic graph of n nodes numbered from 0 to n − 1. This is represented by a 2D array edges of length m, where edges[i] = [ui, vi, costi] indicates a one‑way communication from node ui to node vi with a recovery cost of costi. Some nodes may be offline. You are given a boolean array online where online[i] = true means node i is online. Nodes 0 and n − 1 are always online. A path from 0 to n − 1 is valid if: All intermediate nodes on the path are online. The total recovery cost of all edges on the path does not exceed k. For each valid path, define its score as the minimum edge‑cost along that path. Return the maximum path score (i.e., the largest minimum-edge cost) among all valid paths. If no valid path exists, return -1.

HardProblem #3919
LeetCode

Network Recovery Pathways

Related Topics
ArrayBinary SearchDynamic ProgrammingGraphTopological SortHeap (Priority Queue)Shortest Path
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Examples

Example 1
INPUT
edges = [[0,1,5],[1,3,10],[0,2,3],[2,3,4]], online = [true,true,true,true], k = 10
OUTPUT
3
EXPLANATION

The graph has two possible routes from node 0 to node 3: Path 0 → 1 → 3 Total cost = 5 + 10 = 15, which exceeds k (15 > 10), so this path is invalid. Path 0 → 2 → 3 Total cost = 3 + 4 = 7 <= k, so this path is valid. The minimum edge‐cost along this path is min(3, 4) = 3. There are no other valid paths. Hence, the maximum among all valid path‐scores is 3.

Example 2
INPUT
edges = [[0,1,7],[1,4,5],[0,2,6],[2,3,6],[3,4,2],[2,4,6]], online = [true,true,true,false,true], k = 12
OUTPUT
6
EXPLANATION

Node 3 is offline, so any path passing through 3 is invalid. Consider the remaining routes from 0 to 4: Path 0 → 1 → 4 Total cost = 7 + 5 = 12 <= k, so this path is valid. The minimum edge‐cost along this path is min(7, 5) = 5. Path 0 → 2 → 3 → 4 Node 3 is offline, so this path is invalid regardless of cost. Path 0 → 2 → 4 Total cost = 6 + 6 = 12 <= k, so this path is valid. The minimum edge‐cost along this path is min(6, 6) = 6. Among the two valid paths, their scores are 5 and 6. Therefore, the answer is 6.

Constraints

n == online.length
2 <= n <= 5 * 104
0 <= m == edges.length <= min(105, n * (n - 1) / 2)

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