<p>There is an undirected weighted connected graph. You are given a positive integer <code>n</code> which denotes that the graph has <code>n</code> nodes labeled from <code>1</code> to <code>n</code>, and an array <code>edges</code> where each <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, weight<sub>i</sub>]</code> denotes that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with weight equal to <code>weight<sub>i</sub></code>.</p>

<p>A path from node <code>start</code> to node <code>end</code> is a sequence of nodes <code>[z<sub>0</sub>, z<sub>1</sub>,<sub> </sub>z<sub>2</sub>, ..., z<sub>k</sub>]</code> such that <code>z<sub>0 </sub>= start</code> and <code>z<sub>k</sub> = end</code> and there is an edge between <code>z<sub>i</sub></code> and <code>z<sub>i+1</sub></code> where <code>0 &lt;= i &lt;= k-1</code>.</p>

<p>The distance of a path is the sum of the weights on the edges of the path. Let <code>distanceToLastNode(x)</code> denote the shortest distance of a path between node <code>n</code> and node <code>x</code>. A <strong>restricted path</strong> is a path that also satisfies that <code>distanceToLastNode(z<sub>i</sub>) &gt; distanceToLastNode(z<sub>i+1</sub>)</code> where <code>0 &lt;= i &lt;= k-1</code>.</p>

<p>Return <em>the number of restricted paths from node</em> <code>1</code> <em>to node</em> <code>n</code>. Since that number may be too 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/2021/02/17/restricted_paths_ex1.png" style="width: 351px; height: 341px;" />
<pre>
<strong>Input:</strong> n = 5, edges = [[1,2,3],[1,3,3],[2,3,1],[1,4,2],[5,2,2],[3,5,1],[5,4,10]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Each circle contains the node number in black and its <code>distanceToLastNode value in blue. </code>The three restricted paths are:
1) 1 --&gt; 2 --&gt; 5
2) 1 --&gt; 2 --&gt; 3 --&gt; 5
3) 1 --&gt; 3 --&gt; 5
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/17/restricted_paths_ex22.png" style="width: 356px; height: 401px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[1,3,1],[4,1,2],[7,3,4],[2,5,3],[5,6,1],[6,7,2],[7,5,3],[2,6,4]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Each circle contains the node number in black and its <code>distanceToLastNode value in blue. </code>The only restricted path is 1 --&gt; 3 --&gt; 7.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>n - 1 &lt;= edges.length &lt;= 4 * 10<sup>4</sup></code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i </sub>!= v<sub>i</sub></code></li>
	<li><code>1 &lt;= weight<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
	<li>There is at most one edge between any two nodes.</li>
	<li>There is at least one path between any two nodes.</li>
</ul>

import heapq
from collections import defaultdict

class Solution:
    def countRestrictedPaths(self, n: int, edges: List[List[int]]) -> int:
        # Step 1: Build the graph
        graph = defaultdict(list)
        for u, v, w in edges:
            graph[u].append((v, w))
            graph[v].append((u, w))
        
        # Step 2: Dijkstra's algorithm to find shortest distances from node n
        distances = self.dijkstra(n, graph)
        
        # Step 3: Dynamic Programming to count the number of restricted paths
        dp = {}
        
        def dfs(node):
            if node == n:
                return 1
            if node in dp:
                return dp[node]
            
            total_paths = 0
            for neighbor, _ in graph[node]:
                if distances[node] > distances[neighbor]:
                    total_paths += dfs(neighbor)
            
            dp[node] = total_paths
            return total_paths
        
        return dfs(1) % (10**9 + 7)
    
    def dijkstra(self, start, graph):
        # Initialize distances with infinity
        distances = {node: float('inf') for node in graph}
        distances[start] = 0
        
        # Priority queue to select the node with the smallest distance
        queue = [(0, start)]
        
        while queue:
            current_distance, current_node = heapq.heappop(queue)
            
            # If the distance is greater than the currently known distance, skip
            if current_distance > distances[current_node]:
                continue
            
            # Relaxation step
            for neighbor, weight in graph[current_node]:
                distance = current_distance + weight
                if distance < distances[neighbor]:
                    distances[neighbor] = distance
                    heapq.heappush(queue, (distance, neighbor))
        
        return distances

# Example usage:
solution = Solution()
n = 5
edges = [[1,2,3],[1,3,3],[2,3,1],[1,4,2],[5,2,2],[3,5,1],[5,4,10]]
print(solution.countRestrictedPaths(n, edges))  # Output: 3
1592