#2532

Hard

auto_awesomeArray plus Heap (Priority Queue)

LeetCode Problem Workspace

Time to Cross a Bridge

Time to Cross a Bridge involves simulating worker movements using arrays and heaps to determine when the last worker crosses the bridge.

array heap priority queue simulation

Problem Statement

You are tasked with moving n boxes from an old warehouse on the right side to a new warehouse on the left. There are k workers available, each with their own movement times for crossing the bridge, picking, and putting boxes in the respective warehouses. The movement for each worker is described by a 2D array time where each element [righti, picki, lefti, puti] represents the time spent for crossing the bridge to the right, picking up a box, crossing the bridge to the left, and placing the box in the left warehouse.

Initially, all workers are at the left side of the bridge. Each worker can start by crossing the bridge to the right, picking a box, crossing back to the left, and then placing the box in the new warehouse. The goal is to simulate this process and determine the exact time at which the last worker finishes crossing the bridge to the left side.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

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: See original problem statement.

Output: See original problem statement.

Example details omitted.

Constraints

1 <= n, k <= 104
time.length == k
time[i].length == 4
1 <= lefti, picki, righti, puti <= 1000

Solution Approach

Simulate Worker Movement

Start by simulating the movements of each worker across the bridge. Maintain a priority queue (heap) to keep track of when each worker is available to start their next task. The key here is to determine the precise timing of each worker's tasks and how they interact when crossing the bridge.

Use a Heap to Optimize Worker Scheduling

By using a heap (priority queue), we can manage the tasks of multiple workers efficiently. The heap allows us to quickly select the worker who will be available next, minimizing idle time and ensuring that tasks are completed in the shortest possible time.

Track the Time of the Last Worker

After simulating the process, track the time at which the last worker finishes their task. This is the key result, as it represents the time when the last box is successfully placed in the new warehouse.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the number of workers k and the operations related to the heap. In the worst case, each worker will require multiple heap insertions and deletions, leading to a time complexity of O(k log k). The space complexity is O(k) as we maintain a heap of size k during the simulation.

What Interviewers Usually Probe

Test the candidate's ability to simulate processes efficiently.
Evaluate how well the candidate optimizes worker task management using heaps.
Gauge the candidate's understanding of time complexity in relation to heap operations.

Common Pitfalls or Variants

Common pitfalls

Failing to properly simulate each worker's task time, leading to incorrect timing calculations.
Not utilizing the heap structure correctly to optimize the worker scheduling process.
Overcomplicating the solution instead of using a simple simulation with an efficient heap management.

Follow-up variants

Adding constraints where some workers might have different efficiencies, requiring more advanced scheduling logic.
Allowing workers to perform additional tasks, such as multiple box pick-ups or drops, complicating the simulation process.
Adjusting the time for various tasks, like allowing for different crossing speeds for each worker.

FAQ

What is the main strategy to solve the Time to Cross a Bridge problem?

The main strategy involves simulating the movement of workers using an array for task tracking and a heap to optimize task scheduling.

How does using a heap help with the solution?

A heap efficiently manages the scheduling of workers, ensuring the next available worker is selected with minimal delay and optimizing the process.

What is the time complexity of this problem?

The time complexity is O(k log k), where k is the number of workers, due to heap insertions and deletions during the simulation.

What are some common mistakes to avoid when solving this problem?

Common mistakes include not simulating worker tasks correctly, using inefficient data structures, or overcomplicating the problem when a simple simulation with heaps suffices.

Can the problem be extended or made more complex?

Yes, the problem can be extended by adding more complex worker efficiencies, multiple tasks, or different crossing speeds, requiring advanced simulation techniques.

terminal

Solution

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.

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class Solution:
    def findCrossingTime(self, n: int, k: int, time: List[List[int]]) -> int:
        time.sort(key=lambda x: x[0] + x[2])
        cur = 0
        wait_in_left, wait_in_right = [], []
        work_in_left, work_in_right = [], []
        for i in range(k):
            heappush(wait_in_left, -i)
        while 1:
            while work_in_left:
                t, i = work_in_left[0]
                if t > cur:
                    break
                heappop(work_in_left)
                heappush(wait_in_left, -i)
            while work_in_right:
                t, i = work_in_right[0]
                if t > cur:
                    break
                heappop(work_in_right)
                heappush(wait_in_right, -i)
            left_to_go = n > 0 and wait_in_left
            right_to_go = bool(wait_in_right)
            if not left_to_go and not right_to_go:
                nxt = inf
                if work_in_left:
                    nxt = min(nxt, work_in_left[0][0])
                if work_in_right:
                    nxt = min(nxt, work_in_right[0][0])
                cur = nxt
                continue
            if right_to_go:
                i = -heappop(wait_in_right)
                cur += time[i][2]
                if n == 0 and not wait_in_right and not work_in_right:
                    return cur
                heappush(work_in_left, (cur + time[i][3], i))
            else:
                i = -heappop(wait_in_left)
                cur += time[i][0]
                n -= 1
                heappush(work_in_right, (cur + time[i][1], i))

Continue Practicing

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