Network Delay Time
<p>You are given a network of <code>n</code> nodes, labeled from <code>1</code> to <code>n</code>. You are also given <code>times</code>, a list of travel times as directed edges <code>times[i] = (u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>)</code>, where <code>u<sub>i</sub></code> is the source node, <code>v<sub>i</sub></code> is the target node, and <code>w<sub>i</sub></code> is the time it takes for a signal to travel from source to target.</p>

<p>We will send a signal from a given node <code>k</code>. Return <em>the <strong>minimum</strong> time it takes for all the</em> <code>n</code> <em>nodes to receive the signal</em>. If it is impossible for all the <code>n</code> nodes to receive the signal, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0743.Network%20Delay%20Time/images/931_example_1.png" style="width: 217px; height: 239px;" />
<pre>
<strong>Input:</strong> times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 1
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 2
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= times.length &lt;= 6000</code></li>
	<li><code>times[i].length == 3</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>0 &lt;= w<sub>i</sub> &lt;= 100</code></li>
	<li>All the pairs <code>(u<sub>i</sub>, v<sub>i</sub>)</code> are <strong>unique</strong>. (i.e., no multiple edges.)</li>
</ul>
Depth-First Search; Breadth-First Search; Graph; Shortest Path; Heap (Priority Queue)
function networkDelayTime(times: number[][], n: number, k: number): number {
    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(Infinity));
    for (const [u, v, w] of times) {
        g[u - 1][v - 1] = w;
    }
    const dist: number[] = Array(n).fill(Infinity);
    dist[k - 1] = 0;
    const vis: boolean[] = Array(n).fill(false);
    for (let i = 0; i < n; ++i) {
        let t = -1;
        for (let j = 0; j < n; ++j) {
            if (!vis[j] && (t === -1 || dist[j] < dist[t])) {
                t = j;
            }
        }
        vis[t] = true;
        for (let j = 0; j < n; ++j) {
            dist[j] = Math.min(dist[j], dist[t] + g[t][j]);
        }
    }
    const ans = Math.max(...dist);
    return ans === Infinity ? -1 : ans;
}
