Explain what is DFS (Depth First Search) algorithm for a Graph and how does it work?

Depth-First Search (DFS) is a fundamental algorithm used for traversing or searching tree or graph data structures. The algorithm starts at a specified root node and explores as far as possible along each branch before backtracking. This method is particularly useful for tasks that require exhaustive searches of all nodes, such as solving mazes, detecting cycles, and finding connected components in a graph.

How DFS Works

DFS can be implemented both recursively and iteratively. Here is a step-by-step explanation of how DFS works:

1. Initialization:

   • Start by marking all the vertices as not visited.
   • Use a stack to keep track of the vertices to be visited next. In the recursive implementation, the call stack is used implicitly.

2. Starting Point:

   • Begin the traversal from a specified starting vertex (root node).

3. Traversal:

   • Mark the current vertex as visited.
   • For each adjacent vertex of the current vertex, if it has not been visited, recursively apply DFS to that vertex.
   • If all adjacent vertices have been visited, backtrack to the previous vertex and continue the process.

4. Backtracking:

   • When a vertex has no unvisited adjacent vertices, backtrack to the previous vertex in the stack and continue the traversal from there.

Pseudocode for DFS

Here is the pseudocode for the recursive implementation of DFS:

def DFS(graph, vertex, visited=None):
    if visited is None:
        visited = set()
    visited.add(vertex)
    print(vertex)  # Process the vertex (e.g., print it)
    for neighbor in graph[vertex]:
        if neighbor not in visited:
            DFS(graph, neighbor, visited)
    return visited

Example

Consider the following undirected graph:

0 -- 1 -- 3
|    |
2    4

The adjacency list representation of this graph is:

graph = {
    '0': ['1', '2'],
    '1': ['0', '3', '4'],
    '2': ['0'],
    '3': ['1'],
    '4': ['1']
}

Running DFS starting from vertex '0' would visit the vertices in the following order: 0, 1, 3, 4, 2.

Complexity

• Time Complexity: $$O(V + E)$$, where $$V$$ is the number of vertices and $$E$...