<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:
vector<pair<int, int>> finder(vector<vector<int>> &flights, int k)
{
    vector<pair<int, int>> a;
    for (int i = 0; i < flights.size(); i++)
    {
        if (flights[i][0] == k)
        {
            a.push_back({flights[i][1], flights[i][2]});
        }
    }
    return a;
}

int networkDelayTime(vector<vector<int>> &times, int n, int k)
{
    // Priority queue to store {total_time, node}
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
    vector<int> visitedArray(n + 1, INT_MAX);  // 1-based indexing

    visitedArray[k] = 0;
    pq.push({0, k});  // {time, node}

    while (!pq.empty())
    {
        auto z = pq.top();
        pq.pop();

        int totalPrice = z.first;
        int element = z.second;

        vector<pair<int, int>> elements = finder(times, element);

        for (int i = 0; i < elements.size(); i++)
        {
            int nextElement = elements[i].first;
            int nextPrice = elements[i].second;
            
            if (visitedArray[nextElement] > nextPrice + totalPrice)
            {
                visitedArray[nextElement] = nextPrice + totalPrice;
                pq.push({nextPrice + totalPrice, nextElement});
            }
        }
    }

    // Find the maximum time it takes for the signal to reach all nodes
    int maxTime = 0;
    for (int i = 1; i <= n; i++)  // 1-based, nodes from 1 to n
    {
        if (visitedArray[i] == INT_MAX) return -1;  // If there's a node we can't reach
        maxTime = max(maxTime, visitedArray[i]);
    }

    return maxTime;
}
};
51959