Introduction

In this blog post, we will explore how to solve the LeetCode problem "3218" 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 is an m x n cake that needs to be cut into 1 x 1 pieces. You are given integers m, n, and two arrays: horizontalCut of size m - 1, where horizontalCut[i] represents the cost to cut along the horizontal line i. verticalCut of size n - 1, where verticalCut[j] represents the cost to cut along the vertical line j. In one operation, you can choose any piece of cake that is not yet a 1 x 1 square and perform one of the following cuts: Cut along a horizontal line i at a cost of horizontalCut[i]. Cut along a vertical line j at a cost of verticalCut[j]. After the cut, the piece of cake is divided into two distinct pieces. The cost of a cut depends only on the initial cost of the line and does not change. Return the minimum total cost to cut the entire cake into 1 x 1 pieces. Example 1: Input: m = 3, n = 2, horizontalCut = [1,3], verticalCut = [5] Output: 13 Explanation: Perform a cut on the vertical line 0 with cost 5, current total cost is 5. Perform a cut on the horizontal line 0 on 3 x 1 subgrid with cost 1. Perform a cut on the horizontal line 0 on 3 x 1 subgrid with cost 1. Perform a cut on the horizontal line 1 on 2 x 1 subgrid with cost 3. Perform a cut on the horizontal line 1 on 2 x 1 subgrid with cost 3. The total cost is 5 + 1 + 1 + 3 + 3 = 13. Example 2: Input: m = 2, n = 2, horizontalCut = [7], verticalCut = [4] Output: 15 Explanation: Perform a cut on the horizontal line 0 with cost 7. Perform a cut on the vertical line 0 on 1 x 2 subgrid with cost 4. Perform a cut on the vertical line 0 on 1 x 2 subgrid with cost 4. The total cost is 7 + 4 + 4 = 15. Constraints: 1 <= m, n <= 20 horizontalCut.length == m - 1 verticalCut.length == n - 1 1 <= horizontalCut[i], verticalCut[i] <= 103

Explanation

Here's the breakdown of the solution:

• Greedy Approach: The optimal strategy is to always choose the most expensive cut first. This maximizes the reduction in the number of pieces, thus minimizing the overall cost.
• Sorting: Sort both horizontalCut and verticalCut arrays in descending order. This allows us to iteratively pick the most expensive cuts.

• Counting Pieces: Keep track of the number of horizontal and vertical pieces formed after each cut. The cost of each cut is multiplied by the number of pieces along the other dimension because it applies to all pieces in that direction.

• Time Complexity: O(m log m + n log n) due to sorting. Space Complexity: O(1) excluding the input arrays, since we modify in place.

Code

    def minCost(m: int, n: int, horizontalCut: list[int], verticalCut: list[int]) -> int:
    """
    Calculates the minimum cost to cut a cake into 1x1 pieces.

    Args:
        m: The number of rows in the cake.
        n: The number of columns in the cake.
        horizontalCut: An array of costs to cut along horizontal lines.
        verticalCut: An array of costs to cut along vertical lines.

    Returns:
        The minimum total cost to cut the entire cake into 1 x 1 pieces.
    """
    horizontalCut.sort(reverse=True)
    verticalCut.sort(reverse=True)

    horizontal_pieces = 1
    vertical_pieces = 1
    horizontal_index = 0
    vertical_index = 0
    total_cost = 0

    while horizontal_index < len(horizontalCut) or vertical_index < len(verticalCut):
        if horizontal_index < len(horizontalCut) and (vertical_index == len(verticalCut) or horizontalCut[horizontal_index] >= verticalCut[vertical_index]):
            total_cost += horizontalCut[horizontal_index] * vertical_pieces
            horizontal_pieces += 1
            horizontal_index += 1
        else:
            total_cost += verticalCut[vertical_index] * horizontal_pieces
            vertical_pieces += 1
            vertical_index += 1

    return total_cost