<p>You are given an integer <code>n</code> denoting the number of nodes of a <strong>weighted directed</strong> graph. The nodes are numbered from <code>0</code> to <code>n - 1</code>.</p>

<p>You are also given a 2D integer array <code>edges</code> where <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>, weight<sub>i</sub>]</code> denotes that there exists a <strong>directed</strong> edge from <code>from<sub>i</sub></code> to <code>to<sub>i</sub></code> with weight <code>weight<sub>i</sub></code>.</p>

<p>Lastly, you are given three <strong>distinct</strong> integers <code>src1</code>, <code>src2</code>, and <code>dest</code> denoting three distinct nodes of the graph.</p>

<p>Return <em>the <strong>minimum weight</strong> of a subgraph of the graph such that it is <strong>possible</strong> to reach</em> <code>dest</code> <em>from both</em> <code>src1</code> <em>and</em> <code>src2</code> <em>via a set of edges of this subgraph</em>. In case such a subgraph does not exist, return <code>-1</code>.</p>

<p>A <strong>subgraph</strong> is a graph whose vertices and edges are subsets of the original graph. The <strong>weight</strong> of a subgraph is the sum of weights of its constituent edges.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/02/17/example1drawio.png" style="width: 263px; height: 250px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[0,2,2],[0,5,6],[1,0,3],[1,4,5],[2,1,1],[2,3,3],[2,3,4],[3,4,2],[4,5,1]], src1 = 0, src2 = 1, dest = 5
<strong>Output:</strong> 9
<strong>Explanation:</strong>
The above figure represents the input graph.
The blue edges represent one of the subgraphs that yield the optimal answer.
Note that the subgraph [[1,0,3],[0,5,6]] also yields the optimal answer. It is not possible to get a subgraph with less weight satisfying all the constraints.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/02/17/example2-1drawio.png" style="width: 350px; height: 51px;" />
<pre>
<strong>Input:</strong> n = 3, edges = [[0,1,1],[2,1,1]], src1 = 0, src2 = 1, dest = 2
<strong>Output:</strong> -1
<strong>Explanation:</strong>
The above figure represents the input graph.
It can be seen that there does not exist any path from node 1 to node 2, hence there are no subgraphs satisfying all the constraints.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub>, src1, src2, dest &lt;= n - 1</code></li>
	<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>
	<li><code>src1</code>, <code>src2</code>, and <code>dest</code> are pairwise distinct.</li>
	<li><code>1 &lt;= weight[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def minimumWeight(self, n: int, edges: List[List[int]], src1: int, src2: int, dest: int) -> int:
        # weighted directed graph 
        # edges is src, dst, weight 
        # build graph and it's dual 
        graph = collections.defaultdict(dict)
        dualg = collections.defaultdict(dict)
        for index, edge in enumerate(edges) : 
            src, dst, cost = edge 
            graph[src][dst] = min(cost, graph[src].get(dst, inf))
            dualg[dst][src] = min(cost, dualg[dst].get(src, inf))
        
        # conduct dijkstra's algorithm using priority queue and heaping based on min cost of travel
        def dijkstra(graph_i, src) : 
            distance = {} 
            pq = [(0, src)]
            while pq : 
                cost, node = heapq.heappop(pq)
                if node not in distance : 
                    distance[node] = cost 
                    for neighbor in graph_i[node] : 
                        if neighbor not in distance : 
                            heapq.heappush(pq, (cost + graph_i[node][neighbor], neighbor))
            return distance 
        
        # get src1 distance and src2 distance 
        src1_distance = dijkstra(graph, src1)
        src2_distance = dijkstra(graph, src2)
        # if both of them does not have dest -> auto fail doesn't matter if we can go backward
        if dest not in src1_distance and dest not in src2_distance : 
            return -1 
        else : 
            # otherwise get the dual distance 
            dest_distance = dijkstra(dualg, dest)
            # build travel costs lists 
            travel_costs = set()
            # in order of the distances based on shortest to longest list 
            distance_list = [src1_distance, src2_distance, dest_distance]
            distance_list.sort(key = lambda listing : len(listing))
            # loop over nodes in range n 
            for k in distance_list[0] : 
                cost = distance_list[0][k]
                # if k in second listing 
                if k in distance_list[1] : 
                    cost += distance_list[1][k] 
                    # if k in largest listing (least skipped, do last) 
                    if k in distance_list[2] : 
                        cost += distance_list[2][k]
                        travel_costs.add(cost)
                    else : 
                        continue 
                else : 
                    continue  
            # return min of travel cost if you built it otherwise -1 
            return min(travel_costs) if travel_costs else -1
115000