<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>

import heapq
class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        
        
        # adj = defaultdict(list)
        # for t in times:
        #     adj[t[0]].append((t[2],t[1]))
        # q=[]
        # heapq.heapify(q)
        # dist = [sys.maxsize]*(n)
        # dist[k-1]=0
        # heapq.heappush(q,(0,k))
        # while q:
        #     d,u = heapq.heappop(q)
        #     for weight,v in adj[u]:
        #         if dist[v-1]>dist[u-1]+weight:
        #             dist[v-1]=dist[u-1]+weight
        #             heapq.heappush(q,(dist[v-1],v))
        # for i in range(len(dist)):
        #     if dist[i]==sys.maxsize:
        #         return -1
        # return max(dist)
        
        # this problem is dijkastras algorithm
        # set the distance to all nodes to inf
        # construct edge list
        edges = [[] for _ in range(n) ]
        # print(edges)
        for t in times:
            edges[t[0]-1].append((t[1]-1,t[2])) # (n,d)
        # print(edges)

        dist = [float("inf") for _ in range(n)] # n is the number of nodes
        # visited = [False for _ in range(n)]
        newDist = [(0,k-1)]
        heapq.heapify(newDist)
        while len(newDist)>0:
            # pop from list
            (d,node) = heapq.heappop(newDist)
            d = -d
        
            # visited[node] = True
            dist[node] = d
            # push all the neighbors with this dist
            for nei in edges[node]:
                distance = (d+nei[1])
                if distance < dist[nei[0]]:
                    heapq.heappush(newDist, (-distance,nei[0]))

        maxDist = max(dist)
        if maxDist == float("inf"):
            maxDist = -1

        return maxDist
            
        
        
18000