Note: This document links directly to relevant areas found in the system design topics to avoid duplication. Refer to the linked content for general talking points, tradeoffs, and alternatives.
Gather requirements and scope the problem. Ask questions to clarify use cases and constraints. Discuss assumptions.
Without an interviewer to address clarifying questions, we'll define some use cases and constraints.
Exercise the use of more traditional systems - don't use graph-specific solutions such as GraphQL or a graph database like Neo4j
Clarify with your interviewer if you should run back-of-the-envelope usage calculations.
Handy conversion guide:
Outline a high level design with all important components.
Dive into details for each core component.
Clarify with your interviewer how much code you are expected to write.
Without the constraint of millions of users (vertices) and billions of friend relationships (edges), we could solve this unweighted shortest path task with a general BFS approach:
class Graph(Graph):
def shortest_path(self, source, dest):
if source is None or dest is None:
return None
if source is dest:
return [source.key]
prev_node_keys = self._shortest_path(source, dest)
if prev_node_keys is None:
return None
else:
path_ids = [dest.key]
prev_node_key = prev_node_keys[dest.key]
while prev_node_key is not None:
path_ids.append(prev_node_key)
prev_node_key = prev_node_keys[prev_node_key]
return path_ids[::-1]
def _shortest_path(self, source, dest):
queue = deque()
queue.append(source)
prev_node_keys = {source.key: None}
source.visit_state = State.visited
while queue:
node = queue.popleft()
if node is dest:
return prev_node_keys
prev_node = node
for adj_node in node.adj_nodes.values():
if adj_node.visit_state == State.unvisited:
queue.append(adj_node)
prev_node_keys[adj_node.key] = prev_node.key
adj_node.visit_state = State.visited
return None
We won't be able to fit all users on the same machine, we'll need to shard users across Person Servers and access them with a Lookup Service.
friend_ids
source
and the current user's friend_ids
as the ids for each adjacent_node
adjacent_node
from a given id:
adjacent_node
matching the given id (potential for optimization)Clarify with your interviewer how much code you should be writing.
Note: Error handling is excluded below for simplicity. Ask if you should code proper error handing.
Lookup Service implementation:
class LookupService(object):
def __init__(self):
self.lookup = self._init_lookup() # key: person_id, value: person_server
def _init_lookup(self):
...
def lookup_person_server(self, person_id):
return self.lookup[person_id]
Person Server implementation:
class PersonServer(object):
def __init__(self):
self.people = {} # key: person_id, value: person
def add_person(self, person):
...
def people(self, ids):
results = []
for id in ids:
if id in self.people:
results.append(self.people[id])
return results
Person implementation:
class Person(object):
def __init__(self, id, name, friend_ids):
self.id = id
self.name = name
self.friend_ids = friend_ids
User Graph Service implementation:
class UserGraphService(object):
def __init__(self, lookup_service):
self.lookup_service = lookup_service
def person(self, person_id):
person_server = self.lookup_service.lookup_person_server(person_id)
return person_server.people([person_id])
def shortest_path(self, source_key, dest_key):
if source_key is None or dest_key is None:
return None
if source_key is dest_key:
return [source_key]
prev_node_keys = self._shortest_path(source_key, dest_key)
if prev_node_keys is None:
return None
else:
# Iterate through the path_ids backwards, starting at dest_key
path_ids = [dest_key]
prev_node_key = prev_node_keys[dest_key]
while prev_node_key is not None:
path_ids.append(prev_node_key)
prev_node_key = prev_node_keys[prev_node_key]
# Reverse the list since we iterated backwards
return path_ids[::-1]
def _shortest_path(self, source_key, dest_key, path):
# Use the id to get the Person
source = self.person(source_key)
# Update our bfs queue
queue = deque()
queue.append(source)
# prev_node_keys keeps track of each hop from
# the source_key to the dest_key
prev_node_keys = {source_key: None}
# We'll use visited_ids to keep track of which nodes we've
# visited, which can be different from a typical bfs where
# this can be stored in the node itself
visited_ids = set()
visited_ids.add(source.id)
while queue:
node = queue.popleft()
if node.key is dest_key:
return prev_node_keys
prev_node = node
for friend_id in node.friend_ids:
if friend_id not in visited_ids:
friend_node = self.person(friend_id)
queue.append(friend_node)
prev_node_keys[friend_id] = prev_node.key
visited_ids.add(friend_id)
return None
We'll use a public REST API:
$ curl https://social.com/api/v1/friend_search?person_id=1234
Response:
{
"person_id": "100",
"name": "foo",
"link": "https://social.com/foo",
},
{
"person_id": "53",
"name": "bar",
"link": "https://social.com/bar",
},
{
"person_id": "1234",
"name": "baz",
"link": "https://social.com/baz",
},
For internal communications, we could use Remote Procedure Calls.
Identify and address bottlenecks, given the constraints.
Important: Do not simply jump right into the final design from the initial design!
State you would 1) Benchmark/Load Test, 2) Profile for bottlenecks 3) address bottlenecks while evaluating alternatives and trade-offs, and 4) repeat. See Design a system that scales to millions of users on AWS as a sample on how to iteratively scale the initial design.
It's important to discuss what bottlenecks you might encounter with the initial design and how you might address each of them. For example, what issues are addressed by adding a Load Balancer with multiple Web Servers? CDN? Master-Slave Replicas? What are the alternatives and Trade-Offs for each?
We'll introduce some components to complete the design and to address scalability issues. Internal load balancers are not shown to reduce clutter.
To avoid repeating discussions, refer to the following system design topics for main talking points, tradeoffs, and alternatives:
To address the constraint of 400 average read requests per second (higher at peak), person data can be served from a Memory Cache such as Redis or Memcached to reduce response times and to reduce traffic to downstream services. This could be especially useful for people who do multiple searches in succession and for people who are well-connected. Reading 1 MB sequentially from memory takes about 250 microseconds, while reading from SSD takes 4x and from disk takes 80x longer.1
Below are further optimizations:
Additional topics to dive into, depending on the problem scope and time remaining.
Refer to the security section.
See Latency numbers every programmer should know.