// 802. Find Eventual Safe States #include "leetcode.h" /* * 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. */ class Solution { public: vector<int> eventualSafeNodes(vector<vector<int>> &graph) { int n = graph.size(); vector<vector<int>> g(n); vector<int> dp(n, 0); for (int i = 0; i < n; i++) for (int j = 0; j < graph[i].size(); j++) { g[graph[i][j]].push_back(i); dp[i]++; } queue<int> q; for (int i = 0; i < n; i++) if (!dp[i]) q.push(i); vector<int> ans; while (!q.empty()) { int u = q.front(); q.pop(); ans.push_back(u); for (int v : g[u]) if (!--dp[v]) q.push(v); } sort(ans.begin(), ans.end()); return ans; } }; int main() { Solution obj; vvd(int) g1 = {{1, 2}, {2, 3}, {5}, {0}, {5}, {}, {}}; vvd(int) g2 = {{1, 2, 3, 4}, {1, 2}, {3, 4}, {0, 4}, {}}; for (auto i : obj.eventualSafeNodes(g1)) printf("%d ", i); // expect: 2 4 5 6 printf("\n"); for (auto i : obj.eventualSafeNodes(g2)) printf("%d ", i); // expect: 4 printf("\n"); }