Breadth First Search time complexity analysis

Question

The time complexity to go over each adjacent edge of a vertex is, say, O(N), where N is number of adjacent edges. So, for V numbers of

Answer 1

Considering the following Graph we see how the time complexity is O(|V|+|E|) but not O(V*E).

Adjacency List

V     E
v0:{v1,v2} 
v1:{v3}
v2:{v3}
v3:{}

Operating How BFS Works Step by Step

Step1:

Adjacency lists:

V     E
v0: {v1,v2} mark, enqueue v0
v1: {v3}
v2: {v3}
v3: {}

Step2:

Adjacency lists:

V     E
v0: {v1,v2} dequeue v0;mark, enqueue v1,v2
v1: {v3}
v2: {v3}
v3: {}

Step3:

Adjacency lists:

V     E
v0: {v1,v2}
v1: {v3} dequeue v1; mark,enqueue v3
v2: {v3}
v3: {}

Step4:

Adjacency lists:

V     E
v0: {v1,v2}
v1: {v3}
v2: {v3} dequeue v2, check its adjacency list (v3 already marked)
v3: {}

Step5:

Adjacency lists:

V     E
v0: {v1,v2}
v1: {v3}
v2: {v3}
v3: {} dequeue v3; check its adjacency list

Step6:

Adjacency lists:

V     E
v0: {v1,v2} |E0|=2
v1: {v3}    |E1|=1
v2: {v3}    |E2|=1
v3: {}      |E3|=0

Total number of steps:

|V| + |E0| + |E1| + |E2| +|E3| == |V|+|E|
 4  +  2   +  1   +   1  + 0   ==  4 + 4
                           8   ==  8

Assume an adjacency list representation, V is the number of vertices, E the number of edges.

Each vertex is enqueued and dequeued at most once.

Scanning for all adjacent vertices takes O(|E|) time, since sum of lengths of adjacency lists is |E|.

Hence The Time Complexity of BFS Gives a O(|V|+|E|) time complexity.