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 <= 10^5 -
1 <= edges.length <= min(10^5, n * (n - 1) / 2) -
edges[i].length == 2 -
0 <= fromi, toi < n - All pairs
(fromi, toi)are distinct.
SOLUTION:
class Solution:
def findSmallestSetOfVertices(self, n: int, edges: List[List[int]]) -> List[int]:
indegree = {}
for a, b in edges:
indegree[b] = indegree.get(b, 0) + 1
return [i for i in range(n) if i not in indegree]
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