Given a Directed Acyclic Graph (DAG), having N vertices and M edges, The task is to print all path starting from vertex whose in-degree is zero.
Indegree of a vertex is the total number of incoming edges to a vertex.
Example:
Input: N = 6, edges[] = {{5, 0}, {5, 2}, {4, 0}, {4, 1}, {2, 3}, {3, 1}}
Output: All possible paths:
4 0
4 1
5 0
5 2 3 1
Explaination:
The given graph can be represented as:
There are two vertices whose indegree are zero i.e vertex 5 and 4, after exploring these vertices we got the fallowing path:
4 -> 0
4 -> 1
5 -> 0
5 -> 2 -> 3 -> 1
Input: N = 6, edges[] = {{0, 5}}
Output: All possible paths:
0 5
Explaination:
There will be only one possible path in the graph.
Approach:
- The above problem can be solve by exploring all indegree 0 vertices and printing all vertices in path until a vertex having outdegree 0 is encountered. Follow these steps to print all possible paths:
- Create a boolean array indeg0 which store a true value for all those vertex whose indegree is zero.
- Apply DFS on all those vertex whose indegree is 0.
- Print all path starting from a vertex whose indegree is 0 to a vertex whose outdegrees are zero.
Illustration:
For the above graph:
indeg0[] = {False, False, False, False, True, True}
Since indeg[4] = True, so appling DFS on vertex 4 and printing all path terminating to 0 outdegree vertex are as follow:
4 -> 0
4 -> 1
Also indeg[5] = True, so appling DFS on vertex 5 and printing all path terminating to 0 outdegree vertex are as follow:
5 -> 0
5 -> 2 -> 3 -> 1
Here is the implementation of the above approach:
Python3
# Python program to print all# possible paths in a DAG # Recursive function to print all pathsdef dfs(s): # Append the node in path # and set visited path.append(s) visited[s] = True # Path started with a node # having in-degree 0 and # current node has out-degree 0, # print current path if outdeg0[s] and indeg0[path[0]]: print(*path) # Recursive call to print all paths for node in adj[s]: if not visited[node]: dfs(node) # Remove node from path # and set unvisited path.pop() visited[s] = False def print_all_paths(n): for i in range(n): # for each node with in-degree 0 # print all possible paths if indeg0[i] and adj[i]: path = [] visited = [False] * (n + 1) dfs(i) # Driver codefrom collections import defaultdict n = 6# set all nodes unvisitedvisited = [False] * (n + 1)path = [] # edges = (a, b): a -> bedges = [(5, 0), (5, 2), (2, 3), (4, 0), (4, 1), (3, 1)] # adjacency list for nodesadj = defaultdict(list) # indeg0 and outdeg0 arraysindeg0 = [True]*noutdeg0 = [True]*n for edge in edges: u, v = edge[0], edge[1] # u -> v adj[u].append(v) # set indeg0[v] <- false indeg0[v] = False # set outdeg0[u] <- false outdeg0[u] = False print('All possible paths:')print_all_paths(n) |
All possible paths: 4 0 4 1 5 0 5 2 3 1
Time Complexity: O (N + E)2
Auxiliary Space: O (N)
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