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1557. Minimum Number of Vertices to Reach All Nodes.md

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1557. Minimum Number of Vertices to Reach All Nodes

the second one in Biweekly Contest 33.

Difficulty : Medium

Related Topics : Graph

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:

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:

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

  • mine
    • Java
      • TopologicalSort Runtime: 14 ms, faster than 83.37%, Memory Usage: 142.7 MB, less than 16.80% of Java online submissions 
        //O(N)time
//O(N)space
public List<Integer> findSmallestSetOfVertices(int n, List<List<Integer>> edges) {
    int[] arr = new int[n];
    for(List<Integer> edge : edges){
        arr[edge.get(1)]++;
    }
    List<Integer> res = new ArrayList<>();
    for(int i = 0; i < n; i++){
        if(arr[i] == 0){
            res.add(i);
        }
    }
    return res;
}