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)

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

Space complexity for an adjacency list of an undirected graph having large values of V (vertices) and E (edges) is ___________?Group of answer choicesO(E+V)O(E)O(V*V)O(V)Next

What is the time complexity of detecting a cycle in an undirected graph using DFS?O(V)O(V+E)O(ElogV)O(V^2)

Time complexity to check if an edge exists between two vertices would be ___________?Group of answer choicesO(E)O(V*V)O(V+E)O(1)

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