Program to check we can visit any city from any city or not in Python

Python Server Side Programming Programming

Suppose we have n cities represented as a number in range [0, n) and we also have a list of one-way roads that connects one city to another. We have to check whether we can reach any city from any city.

So, if the input is like n = 3, roads = [[0, 1],[0, 2],[1,0],[1,2],[2,0],[2,1]], then the output will be True, as You can go 0 to 1 and 1 to 0

To solve this, we will follow these steps−

Define a function dfs() . This will take i, visited, g
mark i as visited
for each j in g[i], do
- if j is not visited, then
  - dfs(j, visited, g)
Define a function travel() . This will take g
visited := a new set
dfs(0, visited, g)
return true when size of visited is same as n
From the main method, do the following−
graph := an empty map
rev_graph := an empty map
for each source u, and destination v in roads, do
- insert v at the end of graph[u]
- insert u at the end of rev_graph[v]
return true when travel(graph) and travel(rev_graph) both are true

Let us see the following implementation to get better understanding −

Example

Live Demo

class Solution:
   def solve(self, n, roads):
      from collections import defaultdict
         graph = defaultdict(list)
         rev_graph = defaultdict(list)
   for u, v in roads:
      graph[u].append(v)
      rev_graph[v].append(u)
      def dfs(i, visited, g):
      visited.add(i)
   for j in g[i]:
      if j not in visited:
         dfs(j, visited,g)
         def travel(g):
         visited = set()
         dfs(0, visited, g)
      return len(visited)==n
   return travel(graph) and travel(rev_graph)
ob = Solution()
n = 3
roads = [[0, 1],[0, 2],[1,0],[1,2],[2,0],[2,1]]
print(ob.solve(n, roads))

Input

3, [[0, 1],[0, 2],[1,0],[1,2],[2,0],[2,1]]

Output

True

Arnab Chakraborty

Updated on: 2020-10-05T14:25:52+05:30

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