<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:
    
    class comp
    {
        public:
        bool operator()(vector<int> &a, vector<int> &b)
        {
            if(a[3] > b[3])
                return true;
            else
                return false;
        }
    };
    
    int util(vector<vector<vector<int>>> &g, int src)
    {
        int n = g.size();
        
        priority_queue<vector<int>, vector<vector<int>>, comp> pq;
    
        pq.push({src, -1, 0, 0});
        
        vector<int> vis(n, 0);
        
        int ans = -(int)1e9;
        
        while(pq.size() > 0)
        {
            vector<int> rv = pq.top();
            pq.pop();
            
            int u = rv[0];
            int par = rv[1];
            int w = rv[2];
            int wsf = rv[3];
            
            if(vis[u] == 1)
                continue;
            
            vis[u] = 1;
            ans = max(ans, wsf);
            
            for(vector<int> &t : g[u])
            {
                if(vis[t[0]] == 0)
                {
                    pq.push({t[0], u, t[1], wsf + t[1]});
                }
            }
            
        }
        
        
        for(int i = 1; i < n ; i++)
        {
            if(vis[i] == 0)
                return -1;
        }
    
        return ans;
    
    
    }
    
    int networkDelayTime(vector<vector<int>>& arr, int N, int k) {
        
        int n = N + 1;
        vector<vector<vector<int>>> g(n, vector<vector<int>> ());
        
        for(int i = 0 ; i < arr.size() ; i++)
        {
            int u = arr[i][0];
            int v = arr[i][1];
            int w = arr[i][2];
            
            g[u].push_back({v, w});
            // g[v].push_back({u, w});
        }
        
        return util(g, k);
        
    }
};
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