Leetcode #815: Bus Routes
In this guide, we solve Leetcode #815 Bus Routes in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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 the 0th bus travels in the sequence 1 -> 5 -> 7 -> 1 -> 5 -> 7 -> 1 -> ...
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Breadth-First Search, Array, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def numBusesToDestination(
self, routes: List[List[int]], source: int, target: int
) -> int:
if source == target:
return 0
g = defaultdict(list)
for i, route in enumerate(routes):
for stop in route:
g[stop].append(i)
if source not in g or target not in g:
return -1
q = [(source, 0)]
vis_bus = set()
vis_stop = {source}
for stop, bus_count in q:
if stop == target:
return bus_count
for bus in g[stop]:
if bus not in vis_bus:
vis_bus.add(bus)
for next_stop in routes[bus]:
if next_stop not in vis_stop:
vis_stop.add(next_stop)
q.append((next_stop, bus_count + 1))
return -1
Complexity
The time complexity is , and the space complexity is , where is the total number of stops on all bus routes. The space complexity is , where is the total number of stops on all bus routes.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.