Why does the time complexity of DFS and BFS depend

Question

The site http://web.eecs.utk.edu/~huangj/CS302S04/notes/graph-searching.html describes that when an adjacency list is used then, DFS and BFS have complexity O(V+E), and if an adjacency matrix is used, the complexity is O(V²). Why is this?

Answer 1

You must note that for exploring each vertex time required for exploring it is only equal to c*x where x is the indegree of the vertex.Since we are interested in finding the overall complexity, the overall time would be c1*x1+c2*x2...cnxn for n nodes.Taking max(ci)=d,we see that the overall time is <=d(sum of indegrees of all vertices)=d*2m=O(m).Here we have computed the time for not one vertex, but all the vertices taken together.But the enqueuing operation takes time O(n), so overalll O(n+m).

Answer 2

In both cases, the runtime depends on how long it takes to iterate across the outgoing edges of a given node. With an adjacency list, the runtime is directly proportional to the number of outgoing edges. Since each node is visited once, the cost is the number of nodes plus the number of edges, which is O(m + n). With am adjacency matrix, the time required to find all outgoing edges is O(n) because all n columns in the row for a node must be inspected. Summing up across all n nodes, this works out to O(n²).

Hope this helps!