Sunday, January 19, 2025
HomeGeneralWhy is the time complexity of both DFS and BFS equal to...

Why is the time complexity of both DFS and BFS equal to O (V + E)?

The time complexity of both Depth-First Search (DFS) and Breadth-First Search (BFS) is indeed O (V + E), where V is the number of vertices (or nodes) and E is the number of edges in the graph. Here’s why:

DFS:

1. Visiting vertices: In DFS, each vertex is visited once. This takes O(V) time.
2. Exploring edges: For each vertex, we explore its adjacent edges. Since each edge is explored once, this takes O(E) time.
3. Total time complexity: Therefore, the total time complexity of DFS is O(V) + O(E) = O (V + E).

See also  Which Book Appeals To A Woman?

BFS:

1. Visiting vertices: In BFS, each vertex is also visited once. This takes O(V) time.
2. Exploring edges: For each vertex, we explore its adjacent edges. Again, since each edge is explored once, this takes O(E) time.
3. Queue operations: In BFS, we use a queue to keep track of vertices to visit next. Queue operations (enqueue and dequeue) take constant time, so they don’t affect the overall time complexity.
4. Total time complexity: Therefore, the total time complexity of BFS is also O(V) + O(E) = O (V + E).

See also  How to Recover Deleted Text Messages and Photos

In summary, both DFS and BFS have a time complexity of O (V + E) because they visit each vertex once and explore each edge once. The specific order in which they visit vertices and edges differs, but the overall time complexity remains the same.

RELATED ARTICLES
0 0 votes
Article Rating

Leave a Reply

0 Comments
Oldest
Newest Most Voted
Inline Feedbacks
View all comments
- Advertisment -

Most Popular

Recent Comments

0
Would love your thoughts, please comment.x
()
x