LeetCode in Kotlin
802. Find Eventual Safe States
Medium
There is a directed graph of n nodes with each node labeled from 0 to n - 1. The graph is represented by a 0-indexed 2D integer array graph where graph[i] is an integer array of nodes adjacent to node i, meaning there is an edge from node i to each node in graph[i].
A node is a terminal node if there are no outgoing edges. A node is a safe node if every possible path starting from that node leads to a terminal node (or another safe node).
Return an array containing all the safe nodes of the graph. The answer should be sorted in ascending order.
Example 1:
Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]
Output: [2,4,5,6]
Explanation: The given graph is shown above. Nodes 5 and 6 are terminal nodes as there are no outgoing edges from either of them. Every path starting at nodes 2, 4, 5, and 6 all lead to either node 5 or 6.
Example 2:
Input: graph = [[1,2,3,4],[1,2],[3,4],[0,4],[]]
Output: [4]
Explanation: Only node 4 is a terminal node, and every path starting at node 4 leads to node 4.
Constraints:
n == graph.length1 <= n <= 1040 <= graph[i].length <= n0 <= graph[i][j] <= n - 1graph[i]is sorted in a strictly increasing order.- The graph may contain self-loops.
- The number of edges in the graph will be in the range
[1, 4 * 104].
Solution
class Solution {
fun eventualSafeNodes(graph: Array<IntArray>): List<Int> {
val res: MutableList<Int> = ArrayList()
val vis = IntArray(graph.size)
for (i in graph.indices) {
if (dfs(graph, i, vis)) {
res.add(i)
}
}
return res
}
private fun dfs(graph: Array<IntArray>, src: Int, vis: IntArray): Boolean {
if (vis[src] != 0) {
return vis[src] == 2
}
vis[src] = 1
for (x in graph[src]) {
if (!dfs(graph, x, vis)) {
return false
}
}
vis[src] = 2
return true
}
}