source: https://leetcode.com/problems/minimum-number-of-vertices-to-reach-all-nodes/description/

### 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.

##### 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.

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class Solution { public List<Integer> findSmallestSetOfVertices(int n, List<List<Integer>> edges) { int[] incoming = new int[n]; //count of incoming edges for each node for(List<Integer> edge: edges){ int from = edge.get(0); int to = edge.get(1); incoming[to]++; } List<Integer> list = new ArrayList<>(); //add a node with incoming count 0 to the list for(int i =0;i<incoming.length;i++){ if(incoming[i]==0) list.add(i); } return list; } } |