Number of Restricted Paths From First to Last Node
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1700-1799/1786.Number%20of%20Restricted%20Paths%20From%20First%20to%20Last%20Node/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1700-1799/1786.Number%20of%20Restricted%20Paths%20From%20First%20to%20Last%20Node/images/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>
Graph; Topological Sort; Dynamic Programming; Shortest Path; Heap (Priority Queue)
class Solution {
    private static final int INF = Integer.MAX_VALUE;
    private static final int MOD = (int) 1e9 + 7;
    private List<int[]>[] g;
    private int[] dist;
    private int[] f;
    private int n;

    public int countRestrictedPaths(int n, int[][] edges) {
        this.n = n;
        g = new List[n + 1];
        for (int i = 0; i < g.length; ++i) {
            g[i] = new ArrayList<>();
        }
        for (int[] e : edges) {
            int u = e[0], v = e[1], w = e[2];
            g[u].add(new int[] {v, w});
            g[v].add(new int[] {u, w});
        }
        PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        q.offer(new int[] {0, n});
        dist = new int[n + 1];
        f = new int[n + 1];
        Arrays.fill(dist, INF);
        Arrays.fill(f, -1);
        dist[n] = 0;
        while (!q.isEmpty()) {
            int[] p = q.poll();
            int u = p[1];
            for (int[] ne : g[u]) {
                int v = ne[0], w = ne[1];
                if (dist[v] > dist[u] + w) {
                    dist[v] = dist[u] + w;
                    q.offer(new int[] {dist[v], v});
                }
            }
        }
        return dfs(1);
    }

    private int dfs(int i) {
        if (f[i] != -1) {
            return f[i];
        }
        if (i == n) {
            return 1;
        }
        int ans = 0;
        for (int[] ne : g[i]) {
            int j = ne[0];
            if (dist[i] > dist[j]) {
                ans = (ans + dfs(j)) % MOD;
            }
        }
        f[i] = ans;
        return ans;
    }
}