Description
We have a list of bus routes. Each routes[i] is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->… forever.
We start at bus stop S (initially not on a bus), and we want to go to bus stop T. Travelling by buses only, what is the least number of buses we must take to reach our destination? Return -1 if it is not possible.
Example:
Input:
routes = [[1, 2, 7], [3, 6, 7]]
S = 1
T = 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.
Note:
1 <= routes.length <= 500.1 <= routes[i].length <= 500.0 <= routes[i][j] < 10 ^ 6.
Solutions
1. BFS
# Time: O(nlogn)
# Space: O(n)
class Solution:
def numBusesToDestination(self, routes: List[List[int]], S: int, T: int) -> int:
if S == T:
return 0
stop_board = collections.defaultdict()
for bus, stops in enumerate(routes):
for stop in stops:
if stop not in stop_board:
stop_board[stop] = [bus]
else:
stop_board[stop].append(bus)
queue = collections.deque([S])
visited = set()
res = 0
while queue:
res += 1
size_stops_reach = len(queue)
for _ in range(size_stops_reach):
cur_stop = queue.popleft()
for bus in stop_board[cur_stop]:
if bus in visited:
continue
visited.add(bus)
for stop in routes[bus]:
if stop == T:
return res
queue.append(stop)
return -1
# 45/45 cases passed (392 ms)
# Your runtime beats 92 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (31.6 MB)