<p>You are given an undirected weighted graph of <code>n</code> nodes numbered from 0 to <code>n - 1</code>. The graph consists of <code>m</code> edges represented by a 2D array <code>edges</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> with weight <code>w<sub>i</sub></code>.</p>

<p>Consider all the shortest paths from node 0 to node <code>n - 1</code> in the graph. You need to find a <strong>boolean</strong> array <code>answer</code> where <code>answer[i]</code> is <code>true</code> if the edge <code>edges[i]</code> is part of <strong>at least</strong> one shortest path. Otherwise, <code>answer[i]</code> is <code>false</code>.</p>

<p>Return the array <code>answer</code>.</p>

<p><strong>Note</strong> that the graph may not be connected.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/03/05/graph35drawio-1.png" style="height: 129px; width: 250px;" />
<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 6, edges = [[0,1,4],[0,2,1],[1,3,2],[1,4,3],[1,5,1],[2,3,1],[3,5,3],[4,5,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[true,true,true,false,true,true,true,false]</span></p>

<p><strong>Explanation:</strong></p>

<p>The following are <strong>all</strong> the shortest paths between nodes 0 and 5:</p>

<ul>
	<li>The path <code>0 -&gt; 1 -&gt; 5</code>: The sum of weights is <code>4 + 1 = 5</code>.</li>
	<li>The path <code>0 -&gt; 2 -&gt; 3 -&gt; 5</code>: The sum of weights is <code>1 + 1 + 3 = 5</code>.</li>
	<li>The path <code>0 -&gt; 2 -&gt; 3 -&gt; 1 -&gt; 5</code>: The sum of weights is <code>1 + 1 + 2 + 1 = 5</code>.</li>
</ul>
</div>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/03/05/graphhhh.png" style="width: 185px; height: 136px;" />
<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 4, edges = [[2,0,1],[0,1,1],[0,3,4],[3,2,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[true,false,false,true]</span></p>

<p><strong>Explanation:</strong></p>

<p>There is one shortest path between nodes 0 and 3, which is the path <code>0 -&gt; 2 -&gt; 3</code> with the sum of weights <code>1 + 2 = 3</code>.</p>
</div>

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

<ul>
	<li><code>2 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>m == edges.length</code></li>
	<li><code>1 &lt;= m &lt;= min(5 * 10<sup>4</sup>, n * (n - 1) / 2)</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li><code>1 &lt;= w<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
	<li>There are no repeated edges.</li>
</ul>

import heapq
from collections import deque

class Solution:
    def findAnswer(self, n: int, edges: List[List[int]]) -> List[bool]:
        adj_list = [{} for _ in range(n)] 
        max_weight = 0
        for i, (u, v, w) in enumerate(edges):
            max_weight = max(max_weight, w)
            if v in adj_list[u]:
                if w < adj_list[u][v]:
                    adj_list[u][v] = (w, i)
                    adj_list[v][u] = (w, i)
            else:
                adj_list[u][v] = (w, i)
                adj_list[v][u] = (w, i)

        threshold = max_weight * n + 1
        dists, pars, firsts = [threshold] * n, [set() for _ in range(n)], [False] * n
        src, target = 0, n - 1
        dists[src] = 0
        frontier = [(0, src, -1)]
        target_reached = False

        while frontier:
            dist, node, par = heapq.heappop(frontier)
            if firsts[node]:
                if dists[node] == dist:
                    pars[node].add(par)
    
                continue

            firsts[node] = True
            pars[node].add(par)
            if node == target:
                target_reached = True

            if target_reached:
                continue

            for nb, (nb_dist, edge_idx) in adj_list[node].items():
                new_dist = dist + nb_dist
                if new_dist > dists[nb]:
                    continue

                dists[nb] = new_dist
                heapq.heappush(frontier, (new_dist, nb, node))

        answer = [False] * len(edges)
        frontier = deque([target])
        seen = [False] * n
        seen[target] = True
        while frontier:
            curr = frontier.popleft()
            for par in pars[curr]:
                if par == -1:
                    continue

                edge_idx = adj_list[par][curr][1]
                answer[edge_idx] = True
                if seen[par]:
                    continue

                seen[par] = True
                frontier.append(par)

        return answer
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