Introduction

In this blog post, we will explore how to solve the LeetCode problem "3219" 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 <= 105 horizontalCut.length == m - 1 verticalCut.length == n - 1 1 <= horizontalCut[i], verticalCut[i] <= 103

Explanation

Here's the breakdown of the problem and the solution:

• Greedy Approach: The optimal strategy is to always choose the cut with the highest cost among the available horizontal and vertical cuts. Since each horizontal cut increases the number of vertical pieces by 1, and each vertical cut increases the number of horizontal pieces by 1, we should prioritize the cuts that save us the most in the long run.

• Sorting: Sort both the horizontalCut and verticalCut arrays in descending order. This allows us to efficiently pick the most expensive cuts at each step.

• Iteration: Iterate through the sorted cuts, prioritizing the horizontal or vertical cut based on their costs. Keep track of the number of horizontal and vertical pieces we currently have and use them to multiply the cost of each cut (because that cut is applied to all current pieces along the other dimension).

• Complexity:

  • Runtime: O(m log m + n log n) due to sorting.
  • Storage: O(1) excluding input arrays.

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: A list of costs for horizontal cuts.
        verticalCut: A list of costs for vertical cuts.

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

    h_pieces = 1
    v_pieces = 1
    h_index = 0
    v_index = 0
    total_cost = 0

    while h_index < len(horizontalCut) or v_index < len(verticalCut):
        if h_index < len(horizontalCut) and (v_index == len(verticalCut) or horizontalCut[h_index] >= verticalCut[v_index]):
            total_cost += horizontalCut[h_index] * v_pieces
            h_pieces += 1
            h_index += 1
        else:
            total_cost += verticalCut[v_index] * h_pieces
            v_pieces += 1
            v_index += 1

    return total_cost