LeetCode in Kotlin
1557. Minimum Number of Vertices to Reach All Nodes
Medium
Given a** directed acyclic graph**, with n vertices numbered from 0 to n-1, and an array edges where edges[i] = [fromi, toi] represents a directed edge from node fromi to node toi.
Find the smallest set of vertices from which all nodes in the graph are reachable. It’s guaranteed that a unique solution exists.
Notice that you can return the vertices in any order.
Example 1:
Input: n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]
Output: [0,3]
Explanation: It’s not possible to reach all the nodes from a single vertex. From 0 we can reach [0,1,2,5]. From 3 we can reach [3,4,2,5]. So we output [0,3].
Example 2:
Input: n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]
Output: [0,2,3]
Explanation: Notice that vertices 0, 3 and 2 are not reachable from any other node, so we must include them. Also any of these vertices can reach nodes 1 and 4.
Constraints:
2 <= n <= 10^51 <= edges.length <= min(10^5, n * (n - 1) / 2)edges[i].length == 20 <= fromi, toi < n- All pairs
(fromi, toi)are distinct.
Solution
class Solution {
fun findSmallestSetOfVertices(n: Int, edges: List<List<Int>>): List<Int> {
val indegree = IntArray(n)
for (edge in edges) {
indegree[edge[1]]++
}
val ans: MutableList<Int> = ArrayList()
for (i in indegree.indices) {
if (indegree[i] == 0) {
ans.add(i)
}
}
return ans
}
}