<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>

class Solution:
    def countRestrictedPaths(self, n: int, edges: List[List[int]]) -> int:
        
        dct_nd    = {}
        dist_to_n = {}
        queue     = deque() # from n-node to 1-node
        visited   = set()
        
        # 1 step: create dictionary with nodes and nodes' distances to n

        # create dictionary with format (weight, node_to)
        # heap will automatically sort weight and node_to in ascending order
        for l, r, w in edges:
            dct_nd[l] = dct_nd.get(l, []) + [(w, r)]
            dct_nd[r] = dct_nd.get(r, []) + [(w, l)]
        
        dist_to_n[n] = 0
        queue.append(n)
        visited.add(n)
        
        hpf = dct_nd[n].copy()   # without '.copy()' hpf will be only pointer and dct_nd[n] could change
        heapify(hpf)
        while hpf:
            el_w, el_nd = heappop(hpf)
            if el_nd in visited:   continue
            
            dist_to_n[el_nd] = el_w
            visited.add(el_nd)
            queue.append(el_nd)

            if el_nd == 1:         break  # you don't need to traverse more if you've reached 1-node
                                          # other distances will be more than distance of 1-node
            for (i_w, i_nd) in dct_nd[el_nd]:
                if i_nd not in visited:
                    heappush(hpf, (el_w + i_w, i_nd))
        # step 1: end

        # add to dictionary one more field: number of routes to n
        dist_to_n = {k: [v, 0]   for k, v in dist_to_n.items()}
        dist_to_n[n] = [dist_to_n[n][0], 1]  # for n-node number of routes = 1
        
        # step 2: Dynamic Programming
        visited.clear()
        while queue:
            # start from n and traverse futher and futher
            nd_prv = queue.popleft()
            visited.add(nd_prv)
            for (w_cur, nd_cur) in dct_nd[nd_prv]:
                if nd_cur not in visited           and   \
                   nd_cur     in dist_to_n.keys()  and dist_to_n[nd_cur][0] > dist_to_n[nd_prv][0]:
                    # to current node add number of routes from previous node
                    dist_to_n[nd_cur][1] += dist_to_n[nd_prv][1]
                    # !!! careful !!! you need to add modulo operation (as said in the task)
                    dist_to_n[nd_cur][1]  = int(dist_to_n[nd_cur][1] % (1e9+7))
        # step 2: End
        
        return dist_to_n[1][1] # return number of routes for 1-node
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