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Suppose we have an edge list of a directed graph, there are n nodes and node names are 0 to n- 1. We also have two integer values a and b. We have to check whether there is any node c such that we can go from c to a and also c to b.

So, if the input is like

And a = 2, b = 3, then the output will be True, because here c = 0, so we have route from 0 to 2 and also 0 to 3.

To solve this, we will follow these steps −

- Define a function DFS() . This will take graph, node, visited
- if node is not visited, then
- mark node as visited
- for each x in graph[node], do
- DFS(graph, x, visited)

- From the main method, do the following −
- graph := generate adjacency list from the edge_list
- visited_a, visited_b := two empty sets
- DFS(graph, a, visited_a)
- DFS(graph, b, visited_b)
- ans := a new list from the intersection of visited_b and visited_a
- if ans is not empty, then
- return True

- return False

Let us see the following implementation to get better understanding −

def edge_list_to_graph(edges): s = set() for x, y in edges: s.add(x) s.add(y) s = len(list(s)) graph = [[] for x in range(s)] for x, y in edges: graph[y].append(x) return graph def DFS(graph, node, visited): if node not in visited: visited.add(node) for x in graph[node]: DFS(graph, x, visited) def solve(edges, a, b): graph = edge_list_to_graph(edges) visited_a, visited_b = set(), set() DFS(graph, a, visited_a) DFS(graph, b, visited_b) ans = list(visited_a.intersection(visited_b)) if ans: return True return False ed_list = [(0, 4),(4, 3),(1, 2),(0, 1),(0, 2),(1, 1)] a = 2 b = 3 print(solve(ed_list, a, b))

[(0, 4),(4, 3),(1, 2),(0, 1),(0, 2),(1, 1)], 2, 3

True

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