<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 {
private:
    unordered_map<int, vector<vector<int>>> adj;
    unordered_map<int, int> quickest;
    unordered_set<int> visited;

public:
    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        auto compare = [](pair<int, int>& a, pair<int, int>& b) {
            return a.second > b.second; // Correct comparison for min-heap
        };
        priority_queue<pair<int,int>, vector<pair<int,int>>, decltype(compare)> lowest(compare);

        for (vector<int>& edge : times) {
            adj[edge[0]].push_back(vector<int>{edge[1], edge[2]});
            // adj[edge[1]].push_back(vector<int>{edge[0], edge[2]});
        }

        lowest.push({k, 0});
        for (int i = 1; i <= n; ++i) {
            quickest[i] = INT_MAX;
        }
        
        while(!lowest.empty()) {
            pair<int, int> settled = lowest.top();
            lowest.pop();
            if (visited.count(settled.first)) continue;
            quickest[settled.first] = settled.second;
            for (vector<int>& neighPair : adj[settled.first]) {
                int neigh = neighPair[0];
                int weight = neighPair[1];
                lowest.push({neigh, weight + settled.second});
            }

            visited.insert(settled.first);
        }

        int max_time = 0;
        for(const auto& [key, value] : quickest) {
            max_time = max(max_time, value);
        }

        return max_time == INT_MAX ? -1 : max_time;
    }
};

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