<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

class Solution:
    def countPaths(self, n: int, roads: List[List[int]]) -> int:
        # Create adjacency list
        graph = defaultdict(list)
        for u, v, time in roads:
            graph[u].append((v, time))
            graph[v].append((u, time))
        
        # Initialize distances and ways
        dist = [float('inf')] * n
        dist[0] = 0
        ways = [0] * n
        ways[0] = 1
        
        # Priority queue to select node with minimum time
        pq = [(0, 0)]  # (time, node)
        
        while pq:
            current_time, current_node = heapq.heappop(pq)
            
            # If the current time is greater than the known shortest time, skip
            if current_time > dist[current_node]:
                continue
            
            # Explore neighbors
            for neighbor, time in graph[current_node]:
                new_time = current_time + time
                
                # If new path is shorter, update distance and ways
                if new_time < dist[neighbor]:
                    dist[neighbor] = new_time
                    ways[neighbor] = ways[current_node]
                    heapq.heappush(pq, (new_time, neighbor))
                # If new path is equal to the shortest path, add ways
                elif new_time == dist[neighbor]:
                    ways[neighbor] = (ways[neighbor] + ways[current_node]) % (10**9 + 7)
        
        # Return the number of ways to reach node n-1
        return ways[n-1] % (10**9 + 7)

# Example usage:
solution = Solution()
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]]
print(solution.countPaths(n, roads))  # Output: 4
22400