Is Minimum Cost for Cutting Cake II easy or hard?

Minimum Cost for Cutting Cake II is labeled Hard because identifying the greedy rule is not immediately obvious. Once the insight is recognized, the implementation is straightforward with sorting and simple counters.

Minimum Cost for Cutting Cake II Python/Java solution

Implement the greedy approach by sorting both cost arrays in descending order. Iterate through them while tracking horizontal and vertical segment counts and add cost multiplied by the opposite segment count. The same logic translates directly to Python, Java, C++, C#, JavaScript, and C.

How to solve Minimum Cost for Cutting Cake II in O(n)?

An exact O(n) solution is generally not used because the greedy strategy requires ordering the cuts by cost. Sorting the horizontal and vertical arrays leads to O((m+n) log(m+n)) complexity, which is considered optimal for this problem. After sorting, the remaining computation is a linear pass.

What is the best approach for Minimum Cost for Cutting Cake II?

The optimal solution uses a greedy strategy combined with sorting. Sort horizontal and vertical cut costs in descending order and always apply the largest remaining cut first. Multiply the cost by the number of segments in the opposite direction, then update the segment count. This minimizes how often large costs get multiplied by large segment counts.

Is Minimum Cost for Cutting Cake II asked at Google/Amazon/Meta?

Greedy cost‑minimization problems similar to Minimum Cost for Cutting Cake II appear in interviews at companies like Amazon, Google, and Meta. The pattern of sorting costs and applying the largest operations first is a common greedy interview technique.

What data structure is used in Minimum Cost for Cutting Cake II?

The solution primarily uses arrays along with sorting. Two pointers or indices iterate through the sorted horizontal and vertical cost arrays while counters track how many segments exist in each direction.

What is the time complexity of Minimum Cost for Cutting Cake II?

The optimal greedy solution runs in O((m+n) log(m+n)) time because both cost arrays must be sorted. After sorting, a single linear pass merges the decisions for horizontal and vertical cuts. Space complexity is O(1) aside from the sorting overhead.

Description Hints 3 Solution Video 39:26

3219. Minimum Cost for Cutting Cake II

Hard54.2% acceptance1 company

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

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

The intended solution uses a Greedy approach.

Hint 2

At each step, we will perform a cut on the line with the highest cost.

Hint 3

If you perform a horizontal cut, can you count the contribution that it adds to each row cut that comes afterward?

39:26

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