Today I have done leetcode's July LeetCoding Challenge with cpp.

July LeetCoding Challenge 12

Description

Find Eventual Safe States

There is a directed graph of n nodes with each node labeled from 0 to n - 1. The graph is represented by a 0-indexed 2D integer array graph where graph[i] is an integer array of nodes adjacent to node i, meaning there is an edge from node i to each node in graph[i].

A node is a terminal node if there are no outgoing edges. A node is a safe node if every possible path starting from that node leads to a terminal node (or another safe node).

Return an array containing all the safe nodes of the graph. The answer should be sorted in ascending order.

Example 1:

Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]
Output: [2,4,5,6]
Explanation: The given graph is shown above.
Nodes 5 and 6 are terminal nodes as there are no outgoing edges from either of them.
Every path starting at nodes 2, 4, 5, and 6 all lead to either node 5 or 6.

Example 2:

Input: graph = [[1,2,3,4],[1,2],[3,4],[0,4],[]]
Output: [4]
Explanation:
Only node 4 is a terminal node, and every path starting at node 4 leads to node 4.

Constraints:

• n == graph.length
• 1 <= n <= 104
• 0 <= graph[i].length <= n
• 0 <= graph[i][j] <= n - 1
• graph[i] is sorted in a strictly increasing order.
• The graph may contain self-loops.
• The number of edges in the graph will be in the range [1, 4 * 104].

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
    int len = graph.size();
    vector<vector<int>> reverseGraph(len);
    vector<int> degree(len);
    for(int i = 0; i < len; ++i) {
      for(auto next : graph[i]) {
        reverseGraph[next].push_back(i);
        degree[i] += 1;
      }
    }

    queue<int> q;
    vector<int> answer;
    for(int i = 0; i < len; ++i) {
      if(!degree[i]) {
        q.push(i);
      }
    }

    while(q.size()) {
      int current = q.front();
      q.pop();
      answer.push_back(current);
      for(auto next : reverseGraph[current]) {
        degree[next] -= 1;
        if(!degree[next]) {
          q.push(next);
        }
      }
    }
    sort(answer.begin(), answer.end());
    return answer;
  }
};

// Accepted
// 112/112 cases passed (215 ms)
// Your runtime beats 51.35 % of cpp submissions
// Your memory usage beats 22.44 % of cpp submissions (61.5 MB)