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

Minimum Cost for Cutting Cake II (LeetCode #3219) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m log m + n log n) time complexity and O(1) space complexity. Key insight: Always cut the most expensive line first to minimize total cost.

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

The optimal solution for Minimum Cost for Cutting Cake II runs in O(m log m + n log n) time and uses O(1) space. The complexity is dominated by the sorting steps for both cut arrays, making it efficient for larger inputs.

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

The brute force approach for Minimum Cost for Cutting Cake II 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, which can be very inefficient. We would try to cut the cake in every possible way and sum up the costs, but this quickly becomes unmanageable as the number of cuts increases. The optimal Optimal Solution approach improves this to O(m log m + n log n).

What algorithm or data structure is used to solve Minimum Cost for Cutting Cake II?

Minimum Cost for Cutting Cake II uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

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

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches. Be prepared to optimize your initial solution.

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

Forgetting to sort the cuts before processing. Not considering the number of pieces created when calculating costs.

#3219

Minimum Cost for Cutting Cake II

Hard

ArrayGreedySortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution uses a greedy approach, where we always cut along the line with the highest cost. This is because the cost of cutting increases with the number of pieces created; thus, we want to maximize the cost of each cut by cutting the most expensive lines first.

⚙️

Algorithm

4 steps

1Step 1: Sort the horizontalCut and verticalCut arrays in descending order.
2Step 2: Initialize total cost to 0, and set initial horizontal and vertical piece counts to 1.
3Step 3: Use two pointers to iterate through both sorted arrays, choosing the higher cost cut at each step.
4Step 4: For each cut, update the total cost based on the number of pieces created and increment the respective piece count.

solution.py26 lines

1# Full working Python code
2def minCostOptimal(m, n, horizontalCut, verticalCut):
3    horizontalCut.sort(reverse=True)
4    verticalCut.sort(reverse=True)
5    total_cost = 0
6    h_pieces = 1
7    v_pieces = 1
8    i, j = 0, 0
9    while i < len(horizontalCut) and j < len(verticalCut):
10        if horizontalCut[i] >= verticalCut[j]:
11            total_cost += horizontalCut[i] * v_pieces
12            h_pieces += 1
13            i += 1
14        else:
15            total_cost += verticalCut[j] * h_pieces
16            v_pieces += 1
17            j += 1
18    while i < len(horizontalCut):
19        total_cost += horizontalCut[i] * v_pieces
20        h_pieces += 1
21        i += 1
22    while j < len(verticalCut):
23        total_cost += verticalCut[j] * h_pieces
24        v_pieces += 1
25        j += 1
26    return total_cost

ℹ

Complexity note: The complexity is dominated by the sorting steps for both cut arrays, making it efficient for larger inputs.

1Always cut the most expensive line first to minimize total cost.
2Sorting helps in efficiently determining the next cut.

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