Is Minimum Cost for Cutting Cake I easy or hard?

Minimum Cost for Cutting Cake I is typically classified as Medium difficulty. The implementation is straightforward once you recognize the greedy insight, but identifying why the largest-cost-first strategy minimizes the total cost requires algorithmic intuition.

Minimum Cost for Cutting Cake I Python/Java solution

Python and Java implementations usually sort both cost arrays in descending order and iterate with two pointers. At each step the algorithm selects the larger cut cost, multiplies it by the current number of opposite-direction segments, and updates the segment count. The logic is identical across Python, Java, C++, and JavaScript.

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

Sort the horizontal and vertical cut costs in descending order. Maintain counters for horizontal and vertical segments. At each step choose the larger remaining cost, add its weighted contribution (cost multiplied by current segments in the opposite direction), and update the segment count. This greedy ordering ensures the total cost is minimized with O((m+n) log(m+n)) complexity.

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

The optimal approach is a greedy strategy that always performs the highest-cost cut first. By sorting the horizontal and vertical cut costs in descending order (or using a max heap), you ensure expensive cuts happen when the number of segments is small. This minimizes the multiplication effect that occurs as the cake gets divided. The time complexity is O((m+n) log(m+n)).

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

This problem is a variation of the classic board-cutting greedy problem frequently asked in interviews at companies like Amazon and Google. The key concept—choosing the largest cost first to minimize multiplicative impact—appears in many greedy interview questions.

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

The solution typically uses arrays combined with sorting or a max priority queue. Sorting enables processing costs in descending order, while a priority queue allows dynamically selecting the highest remaining cut. Both implementations rely on greedy decision making.

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

The optimal greedy solution runs in O((m+n) log(m+n)) time due to sorting or priority queue operations. Each cut is processed exactly once while maintaining segment counts. Space complexity ranges from O(1) with sorting to O(m+n) when using a priority queue.

Description Hints 3 Solution Video 39:26

3218. Minimum Cost for Cutting Cake I

Medium58.5% acceptance

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

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Hint 1

The intended solution uses Dynamic Programming.

Hint 2

Let dp[sx][sy][tx][ty] denote the minimum cost to cut the rectangle into 1 x 1 pieces.

Hint 3

Iterate on the row or column on which you will perform the next cut, after the cut, the current rectangle will be decomposed into two sub-rectangles.

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3218. Minimum Cost for Cutting Cake I

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