The running time complexity of DFS (Depth-first search) traversal algorithm is a. O(logV*logE), where V is the number of vertices and E is the number of edges b. O(logV+logE), where V is the number of vertices and E is the number of edges c. O(V+E), where V is the number of vertices and E is the number of edges d. O(V*E), where V is the number of vertices and E is the number of edges

What is the role of Depth-First Search (DFS) in graph traversal?A) Ensures shortest path between nodesB) Visits nodes depth-wise until no more unvisited nodes are leftC) Calculates the average distance between nodesD) Finds the maximum flow in a graph

The time complexity of DFS on an undirected graph using adjacent list representation is: Group of answer choices O left parenthesis vertical line V vertical line plus vertical line E vertical line right parenthesis O left parenthesis vertical line V vertical line right parenthesis O left parenthesis vertical line V vertical line squared right parenthesis O left parenthesis log space V right parenthesis

Give the time and space complexity of the Tree Search version of Depth-First Search (DFS), in terms of the branching factor b and the maximum depth m of the search tree.Write your answer using big-O notation, and if you need to use powers then write them in TeX notation. For example, n squared would be written as n^2. If the algorithm has worst-case complexity n squared, it would be written as O(n^2).

Time Complexity of DFS is? (V – number of vertices, E – number of edges)Group of answer choicesO(V + E)O(E)O(V*E)O(V)

What is depth-first search (DFS)?Question 3Answera.A uniformed combinatorial search algorithm that expands nodes in a random orderb.A uniformed combinatorial search algorithm that expands nodes in order of their depthc.A uniformed combinatorial search algorithm that expands nodes in order of their distance from the rootd.A uniformed combinatorial search algorithm that expands nodes in order of their breadth