Solving the LeetCode 815. Bus Routes Problem (Python): An Optimized Approach
6 minute read
The Bus Routes problem on LeetCode challenges us to find the minimum number of buses required to travel from a source stop to a destination stop. In this article, we’ll explore an optimized approach to solving this problem.
Introduction
The Bus Routes problem on LeetCode challenges us to find the minimum number of buses required to travel from a source stop to a destination stop. In this article, we’ll explore an optimized approach to solving this problem.
Problem Overview
You are given an array routes representing bus routes where routes[i] is a bus route that the ith bus repeats forever.
- For example, if
routes[0] = [1, 5, 7], this means that the0thbus travels in the sequence1 -> 5 -> 7 -> 1 -> 5 -> 7 -> 1 -> ...forever.
You will start at the bus stop source (You are not on any bus initially), and you want to go to the bus stop target. You can travel between bus stops by buses only.
Return the least number of buses you must take to travel from source to target. Return -1 if it is not possible.
Example 1:
Input: routes = [[1,2,7],[3,6,7]], source = 1, target = 6
Output: 2
Explanation: The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.
Example 2:
Input: routes = [[7,12],[4,5,15],[6],[15,19],[9,12,13]], source = 15, target = 12
Output: -1
Problem Explaination
Consider a scenario where buses follow a set of routes, each consisting of multiple stops. The goal is to determine the minimum number of buses needed to reach a destination stop from a given source stop. It’s a classic graph traversal problem that can be efficiently solved using a breadth-first search (BFS) approach.
Initial Approach
In my initial approach, Irepresented the bus routes and stops using a graph. I employed a breadth-first search algorithm to traverse the graph, exploring possible routes from the source stop to the destination stop. The use of defaultdict and deque from the collections module facilitated the implementation.
Initial Implementation
Let’s take a look at the initial implementation:
class Solution:
def numBusesToDestination(self, routes: List[List[int]], S: int, T: int) -> int:
# Create a defaultdict to represent the graph of stops and routes
graph = defaultdict(set)
# Populate the graph with stops and the routes that pass through them
for i, route in enumerate(routes):
for stop in route:
graph[stop].add(i)
# Initialize a variable to keep track of the minimum number of steps
ans = 0
# Initialize a queue for BFS with the source stop
queue = deque([S])
# Create a set to keep track of visited stops
seen_stop = set()
# Create a set to keep track of visited routes
seen_route = set()
# Mark the source stop as visited
seen_stop.add(S)
# Perform BFS until the queue is empty
while queue:
# Process all stops in the current level of the BFS
for _ in range(len(queue)):
# Dequeue the current stop
stop = queue.popleft()
# Check if the current stop is the destination stop
if stop == T:
return ans
# Iterate over the routes that include the current stop
for routeID in graph[stop]:
# Iterate over the stops in the current route
for new_stop in routes[routeID]:
# Check if the next stop has not been visited
if new_stop not in seen_stop:
# Enqueue the next stop and mark it as visited
queue.append(new_stop)
seen_stop.add(new_stop)
# Increment the number of steps for the next level of BFS
ans += 1
# If the destination stop cannot be reached, return -1
return -1
I utilized a graph to represent bus routes and stops. However, despite passing several test cases, the code encountered Time Limit Exceeded errors for certain inputs.
Identifying the Bottleneck
Upon closer inspection, I identified that the primary bottleneck leading to Time Limit Exceeded errors was the usage of a set for each stop to check for visited stops. This redundancy in operations needed addressing for improved efficiency.
Optimization Strategy
To optimize the solution, I adopted a more streamlined approach. By replaced the sets for each stop with a single seen_stops set, reducing redundant operations. Additionally, I introduced a seen_routes set to avoid revisiting the same route. The removal of an unnecessary outer loop further contributed to the optimization.
Optimized Implementation
Let’s explore the optimized version of the code:
class Solution:
def numBusesToDestination(self, routes: List[List[int]], S: int, T: int) -> int:
# Check if the source stop is the same as the destination stop
if S == T:
return 0
# Create a defaultdict to map stops to the routes that pass through them
stop_to_routes = defaultdict(set)
# Populate the stop_to_routes dictionary
# Key: Stop, Value: Set of route indices that include the stop
for i, route in enumerate(routes):
for stop in route:
stop_to_routes[stop].add(i)
# Initialize a queue for BFS with the source stop and the number of steps taken
queue = deque([(S, 0)])
# Create a set to keep track of visited stops
seen_stops = set([S])
# Create a set to keep track of visited routes
seen_routes = set()
# Perform BFS until the queue is empty
while queue:
# Dequeue the current stop and the number of steps taken
current_stop, steps = queue.popleft()
# Check if the current stop is the destination stop
if current_stop == T:
return steps
# Iterate over the routes that include the current stop
for route_id in stop_to_routes[current_stop]:
# Check if the route has not been visited
if route_id not in seen_routes:
# Mark the route as visited
seen_routes.add(route_id)
# Explore the next stops in the route
for next_stop in routes[route_id]:
# Check if the next stop has not been visited
if next_stop not in seen_stops:
# Mark the next stop as visited
seen_stops.add(next_stop)
# Enqueue the next stop and increment the number of steps
queue.append((next_stop, steps + 1))
# If the destination stop cannot be reached, return -1
return -1
Testing and Results
In the testing phase, the optimized solution successfully passed test cases that previously resulted in Time Limit Exceeded errors. This indicates a significant improvement in the efficiency of the code.
Conclusion
In conclusion, optimizing code for efficiency is crucial in solving complex problems like the Bus Routes challenge. By identifying and addressing bottlenecks, we can create more effective solutions. Understanding the problem thoroughly, implementing a strategic approach, and iteratively optimizing the code are key practices for success.