<p>You are given a <strong>directed</strong> graph of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>, where each node has <strong>at most one</strong> outgoing edge.</p>

<p>The graph is represented with a given <strong>0-indexed</strong> array <code>edges</code> of size <code>n</code>, indicating that there is a directed edge from node <code>i</code> to node <code>edges[i]</code>. If there is no outgoing edge from <code>i</code>, then <code>edges[i] == -1</code>.</p>

<p>You are also given two integers <code>node1</code> and <code>node2</code>.</p>

<p>Return <em>the <strong>index</strong> of the node that can be reached from both </em><code>node1</code><em> and </em><code>node2</code><em>, such that the <strong>maximum</strong> between the distance from </em><code>node1</code><em> to that node, and from </em><code>node2</code><em> to that node is <strong>minimized</strong></em>. If there are multiple answers, return the node with the <strong>smallest</strong> index, and if no possible answer exists, return <code>-1</code>.</p>

<p>Note that <code>edges</code> may contain cycles.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-2.png" style="width: 321px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [2,2,3,-1], node1 = 0, node2 = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> The distance from node 0 to node 2 is 1, and the distance from node 1 to node 2 is 1.
The maximum of those two distances is 1. It can be proven that we cannot get a node with a smaller maximum distance than 1, so we return node 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-4.png" style="width: 195px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [1,2,-1], node1 = 0, node2 = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> The distance from node 0 to node 2 is 2, and the distance from node 2 to itself is 0.
The maximum of those two distances is 2. It can be proven that we cannot get a node with a smaller maximum distance than 2, so we return node 2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == edges.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-1 &lt;= edges[i] &lt; n</code></li>
	<li><code>edges[i] != i</code></li>
	<li><code>0 &lt;= node1, node2 &lt; n</code></li>
</ul>

class Disjoint{
public:
    vector<int>parent;
    vector<int>size;

    Disjoint(int node){
        parent.resize(node + 1);
        size.resize(node + 1, 1);
        for(int i = 0; i <= node; i++){
            parent[i] = i;
        }
    }

    int findParent(int node){
        if(node == parent[node]){
            return node;
        }
        return parent[node] = findParent(parent[node]);
    }

    void unionBySize(int u, int v){
        int ulp_u = findParent(u);
        int ulp_v = findParent(v);
        if(ulp_u == ulp_v) return;

        if(size[ulp_u] < size[ulp_v]){
            parent[ulp_u] = ulp_v;
            size[ulp_v] += size[ulp_u];
        }
        else{
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v];
        }
    }
};

class Solution {
public:
    int closestMeetingNode(vector<int>& edges, int node1, int node2) {
        int n = edges.size();
        Disjoint ds(n);

        // Build the disjoint sets based on the graph edges
        for(int i = 0; i < n; i++){
            if(edges[i] != -1) {
                ds.unionBySize(i, edges[i]);
            }
        }

        // If node1 and node2 are not in the same connected component
        if(ds.findParent(node1) != ds.findParent(node2)){
            return -1;
        }

        // BFS to find the minimum meeting node
        vector<int> dist1(n, -1), dist2(n, -1);

        auto bfs = [&](int start, vector<int>& dist) {
            queue<int> q;
            q.push(start);
            dist[start] = 0;

            while(!q.empty()){
                int node = q.front();
                q.pop();
                int neighbor = edges[node];
                if(neighbor != -1 && dist[neighbor] == -1) {
                    dist[neighbor] = dist[node] + 1;
                    q.push(neighbor);
                }
            }
        };

        bfs(node1, dist1);
        bfs(node2, dist2);

        int minDistance = INT_MAX;
        int resultNode = -1;

        for(int i = 0; i < n; i++){
            if(dist1[i] != -1 && dist2[i] != -1){
                int maxDist = max(dist1[i], dist2[i]);
                if(maxDist < minDistance){
                    minDistance = maxDist;
                    resultNode = i;
                } else if(maxDist == minDistance && i < resultNode){
                    resultNode = i;
                }
            }
        }

        return resultNode;
    }
};

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