LeetCode 1786: Number of Restricted Paths From First to Last Node Solution

Master LeetCode problem 1786 (Number of Restricted Paths From First to Last Node), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1786. Number of Restricted Paths From First to Last Node

Medium

Dynamic Programming Graph Topological Sort Heap (Priority Queue)Shortest Path

Problem Explanation

Explanation:

To solve this problem, we can use Dijkstra's algorithm to find the shortest distance from node n to all other nodes. Then, we can use dynamic programming to count the number of restricted paths from node 1 to node n based on the shortest distances computed.

Compute the shortest distances from node n to all other nodes using Dijkstra's algorithm.
Initialize an array dp to store the number of restricted paths from node 1 to each node.
Iterate over the nodes in decreasing order of shortest distance from node n.
For each node u in the iteration, loop through its neighbors v and update dp[u] based on the sum of dp[v] if the distance to v is greater than the distance to u.
Return dp[1] as the number of restricted paths from node 1 to node n.

Time complexity: O(ElogN) where E is the number of edges and N is the number of nodes. Space complexity: O(N)

Solution Code

class Solution {
    public int countRestrictedPaths(int n, int[][] edges) {
        int mod = 1000000007;
        List<int[]>[] graph = new ArrayList[n + 1];
        for (int i = 1; i <= n; i++) {
            graph[i] = new ArrayList<>();
        }
        for (int[] edge : edges) {
            graph[edge[0]].add(new int[] {edge[1], edge[2]});
            graph[edge[1]].add(new int[] {edge[0], edge[2]});
        }
        
        int[] dist = new int[n + 1];
        Arrays.fill(dist, Integer.MAX_VALUE);
        dist[n] = 0;
        
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[1] - b[1]);
        pq.offer(new int[] {n, 0});
        
        while (!pq.isEmpty()) {
            int[] curr = pq.poll();
            int u = curr[0];
            for (int[] neighbor : graph[u]) {
                int v = neighbor[0];
                int weight = neighbor[1];
                if (dist[v] > dist[u] + weight) {
                    dist[v] = dist[u] + weight;
                    pq.offer(new int[] {v, dist[v]});
                }
            }
        }
        
        int[] dp = new int[n + 1];
        dp[n] = 1;
        
        int[] order = new int[n];
        for (int i = 1; i <= n; i++) {
            order[i - 1] = i;
        }
        Arrays.sort(order, (a, b) -> dist[a] - dist[b]);
        
        for (int u : order) {
            for (int[] neighbor : graph[u]) {
                int v = neighbor[0];
                if (dist[v] > dist[u]) {
                    dp[u] = (dp[u] + dp[v]) % mod;
                }
            }
        }
        
        return dp[1];
    }
}

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