17. 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 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.

Solution: (Checking the Indegree of each vertices)

class Solution
{
public:
    vector<int> findSmallestSetOfVertices(int n, vector<vector<int>> &edges)
    {
        vector<int> res;
        vector<int> inDeg(n, 0);
        for (int i = 0; i < edges.size(); i++)
        {
            inDeg[edges[i][1]]++;
        }

        for (int i = 0; i < n; i++)
        {
            if (inDeg[i] == 0)
            {
                res.push_back(i);
            }
        }

        return res;
    }
};