# 882. Reachable Nodes In Subdivided Graph LeetCode Solution

Share:

Reachable Nodes In Subdivided Graph You are given an undirected graph (the “original graph”) with `n` nodes labeled from `0` to `n - 1`. You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge.

The graph is given as a 2D array of `edges` where `edges[i] = [ui, vi, cnti]` indicates that there is an edge between nodes `ui` and `vi` in the original graph, and `cnti` is the total number of new nodes that you will subdivide the edge into. Note that `cnti == 0` means you will not subdivide the edge.

To subdivide the edge `[ui, vi]`, replace it with `(cnti + 1)` new edges and `cnti` new nodes. The new nodes are `x1``x2`, …, `xcnti`, and the new edges are `[ui, x1]``[x1, x2]``[x2, x3]`, …, `[xcnti-1, xcnti]``[xcnti, vi]`.

In this new graph, you want to know how many nodes are reachable from the node `0`, where a node is reachable if the distance is `maxMoves` or less.

Given the original graph and `maxMoves`, return the number of nodes that are reachable from node `0` in the new graph.

Example 1:

```Input: edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3
Output: 13
Explanation: The edge subdivisions are shown in the image above.
The nodes that are reachable are highlighted in yellow.
```

Example 2:

```Input: edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], maxMoves = 10, n = 4
Output: 23
```

Example 3:

```Input: edges = [[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]], maxMoves = 17, n = 5
Output: 1
Explanation: Node 0 is disconnected from the rest of the graph, so only node 0 is reachable.
```

Constraints:

• `0 <= edges.length <= min(n * (n - 1) / 2, 104)`
• `edges[i].length == 3`
• `0 <= ui < vi < n`
• There are no multiple edges in the graph.
• `0 <= cnti <= 104`
• `0 <= maxMoves <= 109`
• `1 <= n <= 3000`

Time: O(n)
Space: O(n)

### C++

`` Will be updated Soon``

### Java

``````
Will be updated Soon``````

### Python

``````  Will be updated Soon
``````

#### Watch Tutorial

Checkout more Solutions here