Hard
You are given an array routes
representing bus routes where routes[i]
is a bus route that the ith
bus repeats forever.
routes[0] = [1, 5, 7]
, this means that the 0th
bus travels in the sequence 1 -> 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
Constraints:
1 <= routes.length <= 500
.1 <= routes[i].length <= 105
routes[i]
are unique.sum(routes[i].length) <= 105
0 <= routes[i][j] < 106
0 <= source, target < 106
import java.util.LinkedList
import java.util.Queue
class Solution {
fun numBusesToDestination(routes: Array<IntArray>, source: Int, target: Int): Int {
if (source == target) {
return 0
}
val targetRoutes: MutableSet<Int> = HashSet()
val queue: Queue<Int> = LinkedList()
val taken = BooleanArray(routes.size)
val graph = buildGraph(routes, source, target, queue, targetRoutes, taken)
if (targetRoutes.isEmpty()) {
return -1
}
var bus = 1
while (queue.isNotEmpty()) {
val size = queue.size
for (i in 0 until size) {
val route = queue.poll()
if (targetRoutes.contains(route)) {
return bus
}
for (nextRoute in graph[route]!!) {
if (!taken[nextRoute]) {
queue.offer(nextRoute)
taken[nextRoute] = true
}
}
}
bus++
}
return -1
}
private fun buildGraph(
routes: Array<IntArray>,
source: Int,
target: Int,
queue: Queue<Int>,
targetRoutes: MutableSet<Int>,
taken: BooleanArray,
): Array<ArrayList<Int>?> {
val len = routes.size
val graph: Array<ArrayList<Int>?> = arrayOfNulls(len)
for (i in 0 until len) {
routes[i].sort()
graph[i] = ArrayList()
var id = routes[i].binarySearch(source)
if (id >= 0) {
queue.offer(i)
taken[i] = true
}
id = routes[i].binarySearch(target)
if (id >= 0) {
targetRoutes.add(i)
}
}
for (i in 0 until len) {
for (j in i + 1 until len) {
if (commonStop(routes[i], routes[j])) {
graph[i]?.add(j)
graph[j]?.add(i)
}
}
}
return graph
}
private fun commonStop(routeA: IntArray, routeB: IntArray): Boolean {
var idA = 0
var idB = 0
while (idA < routeA.size && idB < routeB.size) {
if (routeA[idA] == routeB[idB]) {
return true
} else if (routeA[idA] < routeB[idB]) {
idA++
} else {
idB++
}
}
return false
}
}