# 1976. Number of Ways to Arrive at Destination from collections import defaultdict import heapq """ you are in a city that consists of 'n' intersections numbered from 0 to 'n - 1' with bi directional roads between some intersections. the inputs are generated such that you can reach any intersection from any other intersection and that there is at most one road between any two intersections. you are given an integer 'n' and a 2d integer array 'roads' where 'roads[i] = [u_i, v_i, t_i]' means that there is a road between intersections 'u' and 'v' that takes 't' minutes to travel. you want to know in how many ways you can travel from interseciton 0 to intersection 'n - 1'. """ class Solution(object): def countPaths(self, n, roads): """ :type n: int :type roads: List[List[int]] :rtype: int """ graph = defaultdict(list) for u, v, t in roads: graph[u].append([v, t]) graph[v].append([u, t]) def dijkstra(src): dist, ways = [float("inf")] * n, [0] * n pq = [(0, src)] ways[src] = 1 dist[src] = 0 while pq: d, u = heapq.heappop(pq) if dist[u] < d: continue for v, t in graph[u]: if dist[v] > d + t: dist[v] = d + t ways[v] = ways[u] heapq.heappush(pq, (dist[v], v)) elif dist[v] == d + t: ways[v] = (ways[v] + ways[u]) % (10**9 + 7) return ways[n - 1] return dijkstra(0) if __name__ == "__main__": obj = Solution() print( obj.countPaths( n=7, roads=[ [0, 6, 7], [0, 1, 2], [1, 2, 3], [1, 3, 3], [6, 3, 3], [3, 5, 1], [6, 5, 1], [2, 5, 1], [0, 4, 5], [4, 6, 2], ], ) ) print(obj.countPaths(n=2, roads=[[1, 0, 10]]))