Naive Solution (Brute Force)

A brute-force approach to solve this problem would involve exploring all possible paths from the starting intersection (0) to the destination intersection (n-1) and keeping track of the shortest time taken to reach the destination. Then, we count the number of paths that achieve this shortest time.

1. Explore all paths: Use Depth-First Search (DFS) or Breadth-First Search (BFS) to explore all possible paths from intersection 0 to intersection n-1.
2. Calculate path time: For each path, calculate the total time taken to traverse it.
3. Find shortest time: Determine the shortest time among all paths.
4. Count paths: Count the number of paths that take exactly the shortest time.

This approach is highly inefficient because it explores all possible paths, which can grow exponentially with the number of intersections and roads.

Big O Runtime: O(V^E) where V is number of vertices and E is number of edges, due to exploring all possible paths. Big O Space Usage: O(V), due to the depth of the call stack in DFS or the size of the queue in BFS.

Optimal Solution (Dijkstra with Path Counting)

A more efficient solution involves using Dijkstra's algorithm to find the shortest time from the starting intersection to all other intersections. Simultaneously, we maintain a count of the number of shortest paths to each intersection.

1. Initialization:
   • dist[i] = shortest time from intersection 0 to intersection i (initialized to infinity for all i except 0, which is initialized to 0).
   • ways[i] = number of shortest paths from intersection 0 to intersection i (initialized to 0 for all i except 0, which is initialized to 1).
2. Dijkstra's Algorithm:
   • Use a priority queue to store intersections to visit, prioritized by the shortest time (dist[i]).
   • While the priority queue is not empty:
     • Get the intersection u with the smallest dist[u] from the priority queue.
     • For each neighbor v of u:
       • If dist[u] + time(u, v) < dist[v]:
         • Update dist[v] = dist[u] + time(u, v).
         • Update ways[v] = ways[u] (reset the path count, as we've found a shorter path).
         • Add v to the priority queue.
       • Else if dist[u] + time(u, v) == dist[v]:
         • Update ways[v] = (ways[v] + ways[u]) % (10^9 + 7) (add the number of paths to u to the number of paths to v).
3. Result:
   • ways[n-1] contains the number of shortest paths from intersection 0 to intersection n-1.

Edge Cases:

• If n is 1, return 1.
• Ensure to handle modulo operations correctly to prevent integer overflow.
• If the graph is disconnected, the algorithm will still work, but the result for unreachable nodes will be based on initial infinite distances, so special handling might be needed based on problem interpretation (though the problem statement guarantees connectivity).

Big O Runtime: O(E log V), where V is the number of intersections and E is the number of roads. This is due to Dijkstra's algorithm using a priority queue.

Big O Space Usage: O(V + E), where V is the number of intersections (for dist and ways arrays) and E is the number of roads (for the adjacency list representation of the graph).

import heapq

def count_paths(n: int, roads: list[list[int]]) -> int:
    adj = [[] for _ in range(n)]
    for u, v, time in roads:
        adj[u].append((v, time))
        adj[v].append((u, time))

    dist = [float('inf')] * n
    ways = [0] * n
    dist[0] = 0
    ways[0] = 1
    pq = [(0, 0)]  # (distance, node)
    MOD = 10**9 + 7

    while pq:
        d, u = heapq.heappop(pq)

        if d > dist[u]:
            continue

        for v, time in adj[u]:
            if dist[u] + time < dist[v]:
                dist[v] = dist[u] + time
                ways[v] = ways[u]
                heapq.heappush(pq, (dist[v], v))
            elif dist[u] + time == dist[v]:
                ways[v] = (ways[v] + ways[u]) % MOD

    return ways[n - 1]