<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:
    struct PQElement
    {
        int node;
        int weight;

        // friend bool operator<(PQElement const& lhs, PQElement const& rhs)
        // {
        //     return lhs.weight < rhs.weight;
        // }
        friend bool operator>(PQElement const& lhs, PQElement const& rhs)
        {
            return lhs.weight > rhs.weight;
        }
    };

    int networkDelayTime(vector<vector<int>>& times, int n, int k) 
    {
        set<int> unvisitedNodes;
        for (int i = 1; i <= n; ++i)
        {
            unvisitedNodes.insert(i);
        }

        // Init k min delay table
        vector<int> kMinDelay(n, -1);
        kMinDelay[k-1] = 0;                             // Update table

        std::priority_queue<PQElement, std::vector<PQElement>, std::greater<PQElement>> pq;
        pq.push({k, 0});

        // Sort times into edges map
        std::map<int, vector<vector<int>>> edges;
        for (const auto& time : times)
        {
            if (edges.count(time[0]) == 0)
            {
                edges[time[0]] = vector<vector<int>>();
            }

            edges[time[0]].push_back(time);
        }

        // Loop until priority queue is empty
        while(!pq.empty())
        {
            // Take top element in priority queue and conitune with it
            int currNode = pq.top().node; 
            pq.pop();
            unvisitedNodes.erase(currNode);

            // 2. Go over all edges            
            for (const auto& edge : edges[currNode])
            {
                if (edge[0] == currNode)
                {
                    if (kMinDelay[edge[1] - 1] == -1 || 
                        kMinDelay[edge[1] - 1] >= kMinDelay[edge[0] - 1] +  edge[2])
                    {
                        kMinDelay[edge[1] - 1] = kMinDelay[edge[0] - 1] + edge[2];   // Update table
                    }
                    
                    if (unvisitedNodes.count(edge[1]) == 1)
                    {
                        pq.push({edge[1], edge[2]});
                    }
                }
            }
        }

        if (!unvisitedNodes.empty())
        {
            return -1;
        }

        return *max_element(kMinDelay.begin(), kMinDelay.end());    
    }
};
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