3219. Minimum Cost for Cutting Cake II

Problem Link

Description

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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:

1. Cut along a horizontal line i at a cost of horizontalCut[i].
2. 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

Solution

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Python3

class Solution:
    def minimumCost(self, m: int, n: int, horizontalCut: List[int], verticalCut: List[int]) -> int:
        horizontalCut.sort(reverse = 1)
        verticalCut.sort(reverse = 1)
        hi = vi = 0
        vCut = hCut = 0
        res = 0
 
        while hi < m - 1 or vi < n - 1:
            hCost, vCost = horizontalCut[hi] if hi < m - 1 else -inf, verticalCut[vi] if vi < n - 1 else -inf
 
            if hCost >= vCost:
                res += hCost * (vCut + 1)
                hCut += 1
                hi += 1
            else:
                res += vCost * (hCut + 1)
                vCut += 1
                vi += 1
 
        return res