2532. Time to Cross a Bridge
Description
There are k
workers who want to move n
boxes from the right (old) warehouse to the left (new) warehouse. You are given the two integers n
and k
, and a 2D integer array time
of size k x 4
where time[i] = [righti, picki, lefti, puti]
.
The warehouses are separated by a river and connected by a bridge. Initially, all k
workers are waiting on the left side of the bridge. To move the boxes, the ith
worker can do the following:
- Cross the bridge to the right side in
righti
minutes. - Pick a box from the right warehouse in
picki
minutes. - Cross the bridge to the left side in
lefti
minutes. - Put the box into the left warehouse in
puti
minutes.
The ith
worker is less efficient than the jth
worker if either condition is met:
lefti + righti > leftj + rightj
lefti + righti == leftj + rightj
andi > j
The following rules regulate the movement of the workers through the bridge:
- Only one worker can use the bridge at a time.
- When the bridge is unused prioritize the least efficient worker (who have picked up the box) on the right side to cross. If not, prioritize the least efficient worker on the left side to cross.
- If enough workers have already been dispatched from the left side to pick up all the remaining boxes, no more workers will be sent from the left side.
Return the elapsed minutes at which the last box reaches the left side of the bridge.
Example 1:
Input: n = 1, k = 3, time = [[1,1,2,1],[1,1,3,1],[1,1,4,1]]
Output: 6
Explanation:
From 0 to 1 minutes: worker 2 crosses the bridge to the right. From 1 to 2 minutes: worker 2 picks up a box from the right warehouse. From 2 to 6 minutes: worker 2 crosses the bridge to the left. From 6 to 7 minutes: worker 2 puts a box at the left warehouse. The whole process ends after 7 minutes. We return 6 because the problem asks for the instance of time at which the last worker reaches the left side of the bridge.
Example 2:
Input: n = 3, k = 2, time = [[1,5,1,8],[10,10,10,10]]
Output: 37
Explanation:
The last box reaches the left side at 37 seconds. Notice, how we do not put the last boxes down, as that would take more time, and they are already on the left with the workers.
Constraints:
1 <= n, k <= 104
time.length == k
time[i].length == 4
1 <= lefti, picki, righti, puti <= 1000
Solutions
Solution 1: Priority Queue (Max-Heap and Min-Heap) + Simulation
First, we sort the workers by efficiency in descending order, so the worker with the highest index has the lowest efficiency.
Next, we use four priority queues to simulate the state of the workers:
wait_in_left
: Max-heap, storing the indices of workers currently waiting on the left bank;wait_in_right
: Max-heap, storing the indices of workers currently waiting on the right bank;work_in_left
: Min-heap, storing the time when workers currently working on the left bank finish placing boxes and the indices of the workers;work_in_right
: Min-heap, storing the time when workers currently working on the right bank finish picking up boxes and the indices of the workers.
Initially, all workers are on the left bank, so wait_in_left
stores the indices of all workers. We use the variable cur
to record the current time.
Then, we simulate the entire process. First, we check if any worker in work_in_left
has finished placing boxes at the current time. If so, we move the worker to wait_in_left
and remove the worker from work_in_left
. Similarly, we check if any worker in work_in_right
has finished picking up boxes. If so, we move the worker to wait_in_right
and remove the worker from work_in_right
.
Next, we check if there are any workers waiting on the left bank at the current time, denoted as left_to_go
. At the same time, we check if there are any workers waiting on the right bank, denoted as right_to_go
. If there are no workers waiting to cross the river, we directly update cur
to the next time when a worker finishes placing boxes and continue the simulation.
If right_to_go
is true
, we take a worker from wait_in_right
, update cur
to the current time plus the time it takes for the worker to cross from the right bank to the left bank. If all workers have crossed to the right bank at this point, we directly return cur
as the answer; otherwise, we move the worker to work_in_left
.
If left_to_go
is true
, we take a worker from wait_in_left
, update cur
to the current time plus the time it takes for the worker to cross from the left bank to the right bank, then move the worker to work_in_right
and decrement the number of boxes.
Repeat the above process until the number of boxes is zero. At this point, cur
is the answer.
The time complexity is $O(n \times \log k)$, and the space complexity is $O(k)$. Here, $n$ and $k$ are the number of workers and the number of boxes, respectively.
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