Introduction

In this blog post, we will explore how to solve the LeetCode problem "2532" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 and i > 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

Explanation

• Simulate the process: Use priority queues to track workers waiting on both sides of the bridge and the bridge's availability. Simulate worker movements across the bridge, picking up boxes, and putting them down, updating the time accordingly.
  • Prioritize workers: Prioritize workers based on the inefficiency criteria (left + right), and worker index as tie-breaker. When the bridge is free, prioritize workers on the right side who have picked up boxes, then workers on the left side.
  • Termination Condition: Stop when all boxes have been moved to the left side, and the last worker has crossed to the left side of the bridge.

• Time Complexity: O(n log k), where n is the number of boxes and k is the number of workers. Space Complexity: O(k)

Code

    import heapq

def total_time(n: int, k: int, time: list[list[int]]) -> int:
    """
    Calculates the elapsed time when the last box reaches the left side of the bridge.

    Args:
        n: The number of boxes.
        k: The number of workers.
        time: A 2D array of size k x 4, where time[i] = [righti, picki, lefti, puti].

    Returns:
        The elapsed time at which the last box reaches the left side of the bridge.
    """

    left_wait = []  # (inefficiency, worker_id)
    right_wait = []  # (inefficiency, worker_id)
    left_avail = [True] * k
    right_avail = [True] * k

    for i in range(k):
        heapq.heappush(left_wait, (-(time[i][0] + time[i][2]), -i))  # Negate for max heap

    bridge_free_time = 0
    current_time = 0
    boxes_moved = 0
    workers_sent = 0

    while boxes_moved < n or len(right_wait) > 0 or not all(left_avail):
        # Prioritize workers on the right side
        if bridge_free_time <= current_time and right_wait:
            inefficiency, worker_id = heapq.heappop(right_wait)
            worker_id = -worker_id  # Retrieve original worker_id
            right_avail[worker_id] = True
            bridge_free_time = current_time + time[worker_id][2]
            current_time = bridge_free_time
            left_avail[worker_id] = False

        # Prioritize workers on the left side
        elif bridge_free_time <= current_time and left_wait and workers_sent < n:
            inefficiency, worker_id = heapq.heappop(left_wait)
            worker_id = -worker_id  # Retrieve original worker_id
            left_avail[worker_id] = True
            bridge_free_time = current_time + time[worker_id][0]
            current_time = bridge_free_time
            right_avail[worker_id] = False
            workers_sent += 1

        # Send worker from left to right
        elif bridge_free_time > current_time and (len(right_wait) > 0 or (len(left_wait) > 0 and workers_sent < n)):
          current_time = bridge_free_time
        else:
          break  # No work to do, avoid infinite loop


        # Move worker from right to left
        for i in range(k):
            if not right_avail[i] and bridge_free_time == current_time and boxes_moved < n:
                bridge_free_time += time[i][1]
                boxes_moved += 1
                heapq.heappush(right_wait, (-(time[i][0] + time[i][2]), -i))
                right_avail[i] = True

        #Move worker from left to right
        for i in range(k):
            if not left_avail[i] and bridge_free_time == current_time and boxes_moved < n:
                heapq.heappush(left_wait, (-(time[i][0] + time[i][2]), -i))
                left_avail[i] = True

    return current_time