1557.MinimumNumberofVerticestoReachAllNodes.py

"""
    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:
    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:
    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.
"""
#Difficulty: Medium
#66 / 66 test cases passed.
#Runtime: 1812 ms
#Memory Usage: 52.3 MB

#Runtime: 1812 ms, faster than 20.00% of Python3 online submissions for Minimum Number of Vertices to Reach All Nodes.
#Memory Usage: 52.3 MB, less than 40.00% of Python3 online submissions for Minimum Number of Vertices to Reach All Nodes.

class Solution:
    def findSmallestSetOfVertices(self, n: int, edges: List[List[int]]) -> List[int]:
        start = set()
        end = set()
        for edge in edges:
            start.add(edge[0])
            end.add(edge[1])
        return start.difference(end)