<p>You are in a city that consists of <code>n</code> intersections numbered from <code>0</code> to <code>n - 1</code> with <strong>bi-directional</strong> roads between some intersections. The inputs are generated such that you can reach any intersection from any other intersection and that there is at most one road between any two intersections.</p>

<p>You are given an integer <code>n</code> and a 2D integer array <code>roads</code> where <code>roads[i] = [u<sub>i</sub>, v<sub>i</sub>, time<sub>i</sub>]</code> means that there is a road between intersections <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> that takes <code>time<sub>i</sub></code> minutes to travel. You want to know in how many ways you can travel from intersection <code>0</code> to intersection <code>n - 1</code> in the <strong>shortest amount of time</strong>.</p>

<p>Return <em>the <strong>number of ways</strong> you can arrive at your destination in the <strong>shortest amount of time</strong></em>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/17/graph2.png" style="width: 235px; height: 381px;" />
<pre>
<strong>Input:</strong> n = 7, roads = [[0,6,7],[0,1,2],[1,2,3],[1,3,3],[6,3,3],[3,5,1],[6,5,1],[2,5,1],[0,4,5],[4,6,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The shortest amount of time it takes to go from intersection 0 to intersection 6 is 7 minutes.
The four ways to get there in 7 minutes are:
- 0 ➝ 6
- 0 ➝ 4 ➝ 6
- 0 ➝ 1 ➝ 2 ➝ 5 ➝ 6
- 0 ➝ 1 ➝ 3 ➝ 5 ➝ 6
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, roads = [[1,0,10]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is only one way to go from intersection 0 to intersection 1, and it takes 10 minutes.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 200</code></li>
	<li><code>n - 1 &lt;= roads.length &lt;= n * (n - 1) / 2</code></li>
	<li><code>roads[i].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>1 &lt;= time<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
	<li><code>u<sub>i </sub>!= v<sub>i</sub></code></li>
	<li>There is at most one road connecting any two intersections.</li>
	<li>You can reach any intersection from any other intersection.</li>
</ul>

import heapq
from collections import defaultdict, deque
from typing import List

class Solution:
    def countPaths(self, n: int, roads: List[List[int]]) -> int:
        MOD = 10**9 + 7
        # Create the graph using an adjacency list
        graph = defaultdict(list)
        for start, end, dis in roads:
            graph[start].append((end, dis))
            graph[end].append((start, dis))
        
        # Priority queue to find the shortest paths
        heap = [(0, 0)]  # (distance, node)
        shortest = [float('inf')] * n  # Shortest distance to each node
        shortest[0] = 0
        ways = [0] * n  # Number of ways to reach each node with the shortest distance
        ways[0] = 1
        
        while heap:
            d, node = heapq.heappop(heap)
            
            if d > shortest[n-1]:
                break

            # If the distance to this node is greater than the recorded shortest, skip it
            if d > shortest[node]:
                continue
            
            # Explore neighbors
            for next_node, dist in graph[node]:
                new_dist = d + dist
                
                if new_dist < shortest[next_node]:
                    shortest[next_node] = new_dist
                    ways[next_node] = ways[node]
                    heapq.heappush(heap, (new_dist, next_node))
                elif new_dist == shortest[next_node]:
                    ways[next_node] += ways[node]
                    ways[next_node] %= MOD
        
        # Return the number of ways to reach the target node (n-1) with the shortest distance
        return ways[n - 1]
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