Introduction

In this blog post, we will explore how to solve the LeetCode problem "1687" 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

You have the task of delivering some boxes from storage to their ports using only one ship. However, this ship has a limit on the number of boxes and the total weight that it can carry. You are given an array boxes, where boxes[i] = [ports​​i​, weighti], and three integers portsCount, maxBoxes, and maxWeight. ports​​i is the port where you need to deliver the ith box and weightsi is the weight of the ith box. portsCount is the number of ports. maxBoxes and maxWeight are the respective box and weight limits of the ship. The boxes need to be delivered in the order they are given. The ship will follow these steps: The ship will take some number of boxes from the boxes queue, not violating the maxBoxes and maxWeight constraints. For each loaded box in order, the ship will make a trip to the port the box needs to be delivered to and deliver it. If the ship is already at the correct port, no trip is needed, and the box can immediately be delivered. The ship then makes a return trip to storage to take more boxes from the queue. The ship must end at storage after all the boxes have been delivered. Return the minimum number of trips the ship needs to make to deliver all boxes to their respective ports. Example 1: Input: boxes = [[1,1],[2,1],[1,1]], portsCount = 2, maxBoxes = 3, maxWeight = 3 Output: 4 Explanation: The optimal strategy is as follows: - The ship takes all the boxes in the queue, goes to port 1, then port 2, then port 1 again, then returns to storage. 4 trips. So the total number of trips is 4. Note that the first and third boxes cannot be delivered together because the boxes need to be delivered in order (i.e. the second box needs to be delivered at port 2 before the third box). Example 2: Input: boxes = [[1,2],[3,3],[3,1],[3,1],[2,4]], portsCount = 3, maxBoxes = 3, maxWeight = 6 Output: 6 Explanation: The optimal strategy is as follows: - The ship takes the first box, goes to port 1, then returns to storage. 2 trips. - The ship takes the second, third and fourth boxes, goes to port 3, then returns to storage. 2 trips. - The ship takes the fifth box, goes to port 2, then returns to storage. 2 trips. So the total number of trips is 2 + 2 + 2 = 6. Example 3: Input: boxes = [[1,4],[1,2],[2,1],[2,1],[3,2],[3,4]], portsCount = 3, maxBoxes = 6, maxWeight = 7 Output: 6 Explanation: The optimal strategy is as follows: - The ship takes the first and second boxes, goes to port 1, then returns to storage. 2 trips. - The ship takes the third and fourth boxes, goes to port 2, then returns to storage. 2 trips. - The ship takes the fifth and sixth boxes, goes to port 3, then returns to storage. 2 trips. So the total number of trips is 2 + 2 + 2 = 6. Constraints: 1 <= boxes.length <= 105 1 <= portsCount, maxBoxes, maxWeight <= 105 1 <= ports​​i <= portsCount 1 <= weightsi <= maxWeight

Explanation

Here's the solution to the problem:

• Dynamic Programming: We use dynamic programming to find the minimum number of trips. dp[i] stores the minimum trips needed to deliver the first i boxes.
• Optimization: We iterate through all possible prefixes of boxes that can be taken on a single trip. For each prefix, calculate the number of trips required to deliver those boxes and update the dp array. Key Optimization: precompute the number of port changes for any continuous sub-array of boxes.

• Base Case: dp[0] = 0 (no boxes need 0 trips)

• Runtime Complexity: O(n), where n is the number of boxes.

• Storage Complexity: O(n), due to the dp array, prefix sum arrays and the changes array.

Code

    def box_deliveries(boxes, portsCount, maxBoxes, maxWeight):
    n = len(boxes)
    dp = [float('inf')] * (n + 1)
    dp[0] = 0

    weights = [0] * (n + 1)
    counts = [0] * (n + 1)
    changes = [0] * (n + 1)

    for i in range(1, n + 1):
        weights[i] = weights[i - 1] + boxes[i - 1][1]
        counts[i] = counts[i - 1] + 1
        if i > 1:
            changes[i] = changes[i - 1] + (boxes[i - 1][0] != boxes[i - 2][0])

    for i in range(1, n + 1):
        for j in range(i, 0, -1):
            if counts[i] - counts[j - 1] <= maxBoxes and weights[i] - weights[j - 1] <= maxWeight:
                current_changes = changes[i] - changes[j]
                dp[i] = min(dp[i], dp[j - 1] + current_changes + 2)

    return dp[n]