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
function closestMeetingNode(edges: number[], node1: number, node2: number): number {
    const n = edges.length;
    const g = Array.from({ length: n }, () => []);
    for (let i = 0; i < n; ++i) {
        if (edges[i] != -1) {
            g[i].push(edges[i]);
        }
    }
    const inf = 1 << 30;
    const f = (i: number) => {
        const dist = new Array(n).fill(inf);
        dist[i] = 0;
        const q: number[] = [i];
        while (q.length) {
            i = q.shift();
            for (const j of g[i]) {
                if (dist[j] == inf) {
                    dist[j] = dist[i] + 1;
                    q.push(j);
                }
            }
        }
        return dist;
    };
    const d1 = f(node1);
    const d2 = f(node2);
    let ans = -1;
    let d = inf;
    for (let i = 0; i < n; ++i) {
        const t = Math.max(d1[i], d2[i]);
        if (t < d) {
            d = t;
            ans = i;
        }
    }
    return ans;
}
