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1557. Minimum Number of Vertices to Reach All Nodes

Description

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.

Solutions

  • class Solution {
    public List<Integer> findSmallestSetOfVertices(int n, List<List<Integer>> edges) {
        var cnt = new int[n];
        for (var e : edges) {
            ++cnt[e.get(1)];
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (cnt[i] == 0) {
                ans.add(i);
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> findSmallestSetOfVertices(int n, vector<vector<int>>& edges) {
        vector<int> cnt(n);
        for (auto& e : edges) {
            ++cnt[e[1]];
        }
        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            if (cnt[i] == 0) {
                ans.push_back(i);
            }
        }
        return ans;
    }
};

  • class Solution:
    def findSmallestSetOfVertices(self, n: int, edges: List[List[int]]) -> List[int]:
        cnt = Counter(t for _, t in edges)
        return [i for i in range(n) if cnt[i] == 0]


  • func findSmallestSetOfVertices(n int, edges [][]int) (ans []int) {
	cnt := make([]int, n)
	for _, e := range edges {
		cnt[e[1]]++
	}
	for i, c := range cnt {
		if c == 0 {
			ans = append(ans, i)
		}
	}
	return
}

  • function findSmallestSetOfVertices(n: number, edges: number[][]): number[] {
    const cnt: number[] = new Array(n).fill(0);
    for (const [_, t] of edges) {
        cnt[t]++;
    }
    const ans: number[] = [];
    for (let i = 0; i < n; ++i) {
        if (cnt[i] === 0) {
            ans.push(i);
        }
    }
    return ans;
}


  • impl Solution {
    pub fn find_smallest_set_of_vertices(n: i32, edges: Vec<Vec<i32>>) -> Vec<i32> {
        let mut arr = vec![true; n as usize];
        edges.iter().for_each(|edge| {
            arr[edge[1] as usize] = false;
        });
        arr.iter()
            .enumerate()
            .filter_map(|(i, &v)| if v { Some(i as i32) } else { None })
            .collect()
    }
}

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