Find Closest Node to Given Two Nodes
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2300-2399/2359.Find%20Closest%20Node%20to%20Given%20Two%20Nodes/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2300-2399/2359.Find%20Closest%20Node%20to%20Given%20Two%20Nodes/images/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>
Depth-First Search; Graph
use std::collections::VecDeque;

impl Solution {
    pub fn closest_meeting_node(edges: Vec<i32>, node1: i32, node2: i32) -> i32 {
        let n = edges.len();
        let mut g = vec![Vec::new(); n];
        for i in 0..n {
            if edges[i] != -1 {
                g[i].push(edges[i] as usize);
            }
        }
        let inf = 1 << 30;
        let f = |mut i: usize| -> Vec<i32> {
            let mut dist = vec![inf; n];
            dist[i] = 0;
            let mut q = VecDeque::new();
            q.push_back(i);
            while !q.is_empty() {
                i = q.pop_front().unwrap();
                for &j in &g[i] {
                    if dist[j] == inf {
                        dist[j] = dist[i] + 1;
                        q.push_back(j);
                    }
                }
            }
            dist
        };
        let d1 = f(node1 as usize);
        let d2 = f(node2 as usize);
        let mut ans = -1;
        let mut d = inf;
        for i in 0..n {
            let t = std::cmp::max(d1[i], d2[i]);
            if t < d {
                d = t;
                ans = i as i32;
            }
        }
        ans
    }
}
