Problem of the day
You are given an undirected graph G(V, E), where ‘V’ is the number of vertices and ‘E’ is the number of edges present in the graph and two integers ‘v1’ and ‘v2’ denoting vertices of the graph, find and print the path from ‘v1’ to ‘v2’ (if exists) in reverse order. Print an empty list if there is no path between ‘v1’ and ‘v2’.
Find the path using DFS and print the first path that you encountered.
Note:Vertices are numbered through 0 to V-1.
The first line contains a single integer ‘T’ denoting the number of test cases. Then each test case follows.
The first line of each test case contains two integers ‘V’ and ‘E’ denoting the number of vertices and edges in the graph.
The next ‘E’ lines of the test case contain two space-separated integers ‘a’ and ‘b’ denoting that there exists an edge between ‘a’ and ‘b’.
The last line of the test case contains two space-separated integers ‘v1’ and ‘v2’ denoting the starting vertex and ending vertex.
Output Format :
For each test case, print the path from ‘v1’ to ‘v2’ in reverse order.
Output for each test case will be printed in a separate line.
Note :
You are not required to print anything; it has already been taken care of. Just implement the function and return a list of paths.
If there is no path between the vertices return an empty list.If the path is valid then it will print true else it will print false.
1 <= T <= 10
1 <= V <= 1000
1 <= E <= (V * (V - 1)) / 2
0 <= v1, v2 <= V-1
Time Limit: 1sec
2
4 4
0 1
0 3
1 2
2 3
1 3
6 3
5 3
0 1
3 4
0 3
true
false
In the first test case, there are two paths from 1 to 3. I.e. 1 -> 2 -> 3 or 1 -> 0 -> 3. So you can print any one of them.
In the second test case, there is no path from 0 to 3. Hence return an empty string.
2
2 1
0 1
1 0
5 4
0 1
1 2
2 3
3 4
0 4
true
true
Iterate through the graph and store the parents of every node.
In this approach, iterate through the whole graph and whenever you encounter a new node, update the parent of the new node by the current node.
The steps are as follows :
O(V + E). Where V is the number of nodes and E is the number of edges in the graph.
Since we are implementing Depth-first search which is a linear task, Hence the time complexity is O(V + E).
O(V), Where V is the number of nodes in the graph.
We are using a list par of size V to store the parents of all the nodes, Hence the space complexity is O(V).