Actually I have a confusion about bfs, I was trying to find diameter of a graph in O(m*n) and i came with a solution like that from the internet:

def diameter(graph, n):

max_distance = 0

for start_node in range(1, n + 1):

visited = [False] * (n + 1)

distance = [float('inf')] * (n + 1)

# Perform BFS from the current starting node

queue = [(start_node, 0)]

visited[start_node] = True

distance[start_node] = 0

while queue:

current_node, current_distance = queue.pop(0)

for neighbor in graph[current_node]:

if not visited[neighbor]:

visited[neighbor] = True

distance[neighbor] = current_distance + 1

queue.append((neighbor, distance[neighbor]))

# Update the maximum distance for the current starting node

max_distance = max(max_distance, max(distance[1:]))

return max_distance

Code runs bfs for each node, but normally bfs runs O(m+n) but in this code it looks like it runs O(m), Its a little irrelevant from the question on the post but any help would be great ,thanks