If the adjacent node is already visited: Update → low[node] = min(low[node], tin[adjNode]).

Special case for the starting node:

If parent == -1 and child > 1, then mark this starting node as an articulation point.

Result:

After DFS completes, the answer (hash/boolean array) will store all articulation points.

Important Note: While calculating low[], we do not consider the parent node or already visited nodes, as they themselves may be articulation points and could be removed.

Code

#include <bits/stdc++.h>
using namespace std;

class Solution {
private:
    int timer = 1;

    // DFS to find articulation points
    void dfs(int node, int parent, vector<int> &vis, int tin[], int low[],
             vector<int> &mark, vector<int> adj[]) {

        vis[node] = 1;
        tin[node] = low[node] = timer++; // set discovery and low time
        int child = 0;

        for (auto it : adj[node]) {
            if (it == parent) continue; // skip the edge to parent

            if (!vis[it]) {
                dfs(it, node, vis, tin, low, mark, adj); // recursive DFS
                low[node] = min(low[node], low[it]);     // update low time

                // Check articulation condition (excluding root)
                if (low[it] >= tin[node] && parent != -1) {
                    mark[node] = 1;
                }
                child++;
            }
            else {
          