main.cpp

// Source: https://leetcode.com/problems/minimum-number-of-vertices-to-reach-all-nodes
// Title: Minimum Number of Vertices to Reach All Nodes
// Difficulty: Medium
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// Given a **directed acyclic graph**, with `n` vertices numbered from `0` to `n-1`, and an array `edges` where `edges[i] = [from_i, to_i]` represents a directed edge from node `from_i` to node `to_i`.
//
// 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:**
//
// https://assets.leetcode.com/uploads/2020/07/07/untitled22.png
//
// ```
// 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:**
//
// https://assets.leetcode.com/uploads/2020/07/07/untitled.png
//
// ```
// 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 <= from_i,to_i < n`
// - All pairs `(from_i, to_i)` are distinct.
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <vector>

using namespace std;

// Count Degree
//
// Since the graph is acyclic, just pick the nodes with no in-degree
class Solution {
 public:
  vector<int> findSmallestSetOfVertices(int n, vector<vector<int>>& edges) {
    // Compute in-degrees
    auto inDegrees = vector<int>(n, 0);
    for (const auto& edge : edges) {
      int to = edge[1];
      inDegrees[to]++;
    }

    // Find zero-degree nodes
    auto ans = vector<int>();
    ans.reserve(n);
    for (auto i = 0; i < n; ++i) {
      if (inDegrees[i] == 0) {
        ans.push_back(i);
      }
    }

    return ans;
  }
};