DFS Traversal

Moderate

0/80

Average time to solve is 35m

Contributed by

225 upvotes

Asked in companies

Problem statement

Given an undirected and disconnected graph G(V, E), containing 'V' vertices and 'E' edges, the information about edges is given using 'GRAPH' matrix, where i-th edge is between GRAPH[i][0] and GRAPH[i][1]. print its DFS traversal.

V is the number of vertices present in graph G and vertices are numbered from 0 to V-1. 

E is the number of edges present in graph G.

Note :

The Graph may not be connected i.e there may exist multiple components in a graph.

Detailed explanation ( Input/output format, Notes, Images )

Input Format :

The first line of input will contain two Integers V and E, separated by a single space.

From the second line onwards, the next 'E' lines will denote the undirected edge of the graphs. 

Every edge is defined by two single space-separated integers 'a' and 'b', which signifies an edge between the vertices 'a' and 'b'.

Output Format :

The first line of output will contain the size of the connected components.

For every connected component in the graph, print the vertices of the component in the sorted order of the vertex values separated with a single space.

Print each component in on a different line by making sure that the first vertex of each component is also sorted on the vertex values. 

A component having a smaller first vertex in sorted order will come first.

Constraints :

2 <= V <= 10^3
1 <= E <= (5 * (10^3))

Time Limit: 1sec

Sample Input 1:

Sample Output 1:

2
0 1 2
3 4

Explanation For Sample Input 1:

If we do a DFS traversal from vertex 0 we will get a component with vertices [0, 2, 1]. If we do a DFS traversal from 3 we will get another component with vertices [3, 4]

Hence,  we have two disconnected components so on the first line, print 2. Now, print each component in increasing order. On the first line print 0 1 2 and on the second line, print 3 4.

[0 1 2] comes before [3 4] as the first vertex 0 from the first component is smaller than the first vertex 3 from the second component.

Sample Input 2:

Sample Output 2:

Hint 1

Try to extend the DFS algorithm for this problem. How would you handle the disconnected components? What about performing DFS for each component.

Approaches (1)

Approach 1

DFS For Each Connected Component

Run a loop from 0 to V-1 and if this vertex is not visited do a DFS from this vertex and add all the reachable vertex to a vector/list ‘singleComponent’.
Sort the singleComponent vector/list in increasing order and add it to the answer vector/list which is called ‘components’.
Print the size of the vector/list ‘components’ on the first line.
On each line after the first, print one single component from the vector/list ‘components’.

Time Complexity

O(VLogV + E), Where V is the number of vertices and E is the number of edges in the graph.

The time complexity of a DFS is O(V+E) and we are sorting every component which would be VlogV so overall complexity is O(V+E+VlogV) which is O(VlogV+E).

Space Complexity

O(V+E), Where V is the number of vertices and E is the number of edges in the graph.

To store the graph in the adjacency list.

(75% EXP penalty)

(100% EXP penalty)

DFS Traversal

Console