README_EN.md

1557. Minimum Number of Vertices to Reach All Nodes

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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:

<li><code>2 &lt;= n &lt;= 10^5</code></li>

<li><code>1 &lt;= edges.length &lt;= min(10^5, n * (n - 1) / 2)</code></li>

<li><code>edges[i].length == 2</code></li>

<li><code>0 &lt;= from<sub>i,</sub>&nbsp;to<sub>i</sub> &lt; n</code></li>

<li>All pairs <code>(from<sub>i</sub>, to<sub>i</sub>)</code> are distinct.</li>

Solutions

Python3

class Solution:
    def findSmallestSetOfVertices(self, n: int, edges: List[List[int]]) -> List[int]:
        s = {to for _, to in edges}
        return [i for i in range(n) if i not in s]

Java

class Solution {
    public List<Integer> findSmallestSetOfVertices(int n, List<List<Integer>> edges) {
        Set<Integer> s = new HashSet<>();
        for (List<Integer> e : edges) {
            s.add(e.get(1));
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (!s.contains(i)) {
                ans.add(i);
            }
        }
        return ans;
    }
}

C++

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

Go

func findSmallestSetOfVertices(n int, edges [][]int) []int {
	s := make(map[int]bool)
	for _, e := range edges {
		s[e[1]] = true
	}
	var ans []int
	for i := 0; i < n; i++ {
		if !s[i] {
			ans = append(ans, i)
		}
	}
	return ans
}

TypeScript

function findSmallestSetOfVertices(n: number, edges: number[][]): number[] {
    const arr = new Array(n).fill(true);
    for (const [_, i] of edges) {
        arr[i] = false;
    }
    const res = [];
    arr.forEach((v, i) => {
        if (v) {
            res.push(i);
        }
    });
    return res;
}

Rust

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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