<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://assets.leetcode.com/uploads/2019/05/23/931_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>

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& ad, int V, int S) {
        vector<int> dist(V, 1e8); // Initialize distances to infinity
        dist[S - 1] = 0; // Set distance of the source to 0 (1-based to 0-based index)
        
        // Relax edges V-1 times
        for (int i = 0; i < V - 1; ++i) {
            for (int j = 0; j < ad.size(); ++j) {
                int u = ad[j][0] - 1; // Convert 1-based to 0-based index
                int v = ad[j][1] - 1; // Convert 1-based to 0-based index
                int wt = ad[j][2]; // Weight of the edge
                
                if (dist[u] != 1e8 && dist[u] + wt < dist[v]) {
                    dist[v] = dist[u] + wt;
                }
            }
        }
        
        // Find the maximum distance
        int maxDist = 0;
        for (int i = 0; i < V; ++i) {
            if (dist[i] == 1e8) { // If any node is unreachable, return -1
                return -1;
            }
            maxDist = max(maxDist, dist[i]);
        }
        
        return maxDist; // Maximum time it takes for all nodes to receive the signal
    }
};

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