<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>

from heapq import heappop, heappush
from collections import defaultdict


class Solution:
    def minimumWeight(
        self, n: int, edges: List[List[int]], src1: int, src2: int, dest: int
    ) -> int:
        edge_map = defaultdict(dict) #O(E + NlogE)
        for e in edges:
            if e[1] in edge_map[e[0]]:
                edge_map[e[0]][e[1]] = min(e[2], edge_map[e[0]][e[1]])
            else:
                edge_map[e[0]][e[1]] = e[2]

        distances_from_src1 = self.getDistances(src1, edge_map, set(range(n)))
        distances_from_src2 = self.getDistances(src2, edge_map, set(range(n)))

        edge_map = defaultdict(dict)
        for e in edges:
            if e[0] in edge_map[e[1]]:
                edge_map[e[1]][e[0]] = min(e[2], edge_map[e[1]][e[0]])
            else:
                edge_map[e[1]][e[0]] = e[2]
        distances_to_dest = self.getDistances(
            dest,
            edge_map,
            set(distances_from_src1.keys()) & set(distances_from_src2.keys()),
        )

        best_weight = -1

        for i in distances_to_dest.keys():
            total_weight = (
                distances_from_src1[i] + distances_from_src2[i] + distances_to_dest[i]
            )
            if best_weight == -1:
                best_weight = total_weight
            elif best_weight > total_weight:
                best_weight = total_weight

        return best_weight

    def getDistances(self, src, edge_map, candidates):
        distances = defaultdict()
        queue = [(0, src)]

        while queue:
            dist, fr = heappop(queue)
            if fr in distances or fr not in candidates:
                continue
            distances[fr] = dist
            for to, w in edge_map[fr].items():
                if to not in distances:
                    heappush(queue, (dist + w, to))

        return distances

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