How do you solve Minimum Cost for Cutting Cake I on LeetCode?

Minimum Cost for Cutting Cake I (LeetCode #3218) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m + n) time complexity and O(1) space complexity. Key insight: Choosing cuts that maximize the area being cut minimizes total cost.

What is the brute force solution for Minimum Cost for Cutting Cake I?

The brute force approach for Minimum Cost for Cutting Cake I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves calculating the cost of every possible sequence of cuts until the cake is fully divided into 1x1 pieces. This method is straightforward but inefficient, as it explores all combinations of cuts. The optimal Optimal Solution approach improves this to O(m + n).

What are the interview tips for Minimum Cost for Cutting Cake I?

Always explain your thought process clearly when choosing cuts. Be prepared to discuss the trade-offs of different approaches. Practice similar problems to recognize patterns in cutting or partitioning tasks.

What are common mistakes when solving Minimum Cost for Cutting Cake I?

Forgetting to sort the cut costs before processing. Not correctly updating the number of pieces after each cut.

#3218

Minimum Cost for Cutting Cake I

Medium

ArrayTwo PointersDynamic ProgrammingGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m + n)
Space	O(1)	O(1)

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Intuition

Time O(m + n)Space O(1)

The optimal solution uses a greedy approach, where we always choose the cut that maximizes the area being cut at the minimum cost. By sorting the cut costs and using a two-pointer technique, we can efficiently calculate the total cost.

⚙️

Algorithm

3 steps

1Step 1: Sort both horizontalCut and verticalCut arrays.
2Step 2: Initialize total cost, hCount, and vCount to 1 (since we start with one piece).
3Step 3: Use two pointers to iterate through both cut arrays, adding the cost of the cut multiplied by the current count of pieces in the opposite direction.

solution.py20 lines

1# Full working Python code
2class Solution:
3    def minCost(self, m: int, n: int, horizontalCut: List[int], verticalCut: List[int]) -> int:
4        horizontalCut.sort()
5        verticalCut.sort()
6        total_cost = 0
7        h_count = 1
8        v_count = 1
9        h_index = 0
10        v_index = 0
11        while h_index < len(horizontalCut) or v_index < len(verticalCut):
12            if (h_index < len(horizontalCut) and (v_index >= len(verticalCut) or horizontalCut[h_index] <= verticalCut[v_index])):
13                total_cost += horizontalCut[h_index] * v_count
14                h_count += 1
15                h_index += 1
16            else:
17                total_cost += verticalCut[v_index] * h_count
18                v_count += 1
19                v_index += 1
20        return total_cost

ℹ

Complexity note: The time complexity is linear because we only traverse each cut array once after sorting, making it efficient compared to the brute-force approach.

1Choosing cuts that maximize the area being cut minimizes total cost.
2Sorting the cut costs allows for efficient selection of the next cut.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.