# 1557. Minimum Number of Vertices to Reach All Nodes

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 from i to node to i. 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.

# Solution

Approach 1: find indegrees, then deal with loops.

# Code (Python)

Approach 1:

class Solution:
    def findSmallestSetOfVertices(self, n: int, neighbors: List[List[int]]) -> List[int]:
        unreachables = set([i for i in range(n)])
        zero_indegree_nodes = set([i for i in range(n)])
        edges = {}
        for entry in neighbors:
            if entry[0] not in edges:
                edges[entry[0]] = []
        # step 1: find all vertices with indegree = 0, traverse them and mark up reachable nodes
        result = list(zero_indegree_nodes)
        for node in zero_indegree_nodes:
            self._explore(node, unreachables, edges)
        # step 2: the rests are loops that are not connected -- also traverse and mark up
        while unreachables:
            self._explore(min(unreachables), unreachables, edges)
        return sorted(result)
    def _explore(self, node, not_seen, edges):
        if node not in not_seen:
        if node in edges:
            for neighbor in edges[node]:
                self._explore(neighbor, not_seen, edges)

# Code (C++)

Approach 1:

Approach 2: