Question

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:

• 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

Solution 1

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]

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;
  }
};

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()
  }
}