Minimum Cost for Cutting Cake I

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Greedy Algorithms

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the cut positions and the cake size?
Are the cut positions guaranteed to be within the bounds of the cake (i.e., between 0 and the cake size)?
If there are no cuts to be made, should I return 0?
Are the cut positions provided in any particular order, or can they be unsorted?
What data type represents the cost of a cut, and can it be negative?

Brute Force Solution

Approach

The brute force method for cutting the cake explores every possible combination of cuts to find the cheapest one. It's like trying every single way to slice the cake and calculating the cost for each, then picking the cheapest option. This guarantees finding the minimum cost, but it can take a very long time.

Here's how the algorithm would work step-by-step:

Consider every possible place you could make the first cut in the cake.
For each of those first cut options, consider all the possible places to make the second cut.
Keep doing this until you've made all the required cuts.
For each complete set of cuts, calculate the total cost of those cuts.
Compare the total cost of each set of cuts.
Choose the set of cuts that has the smallest total cost. That's the minimum cost.

Code Implementation

def minimum_cost_cake_cutting_brute_force(cake_length, cuts):
    cuts.sort()
    number_of_cuts = len(cuts)

    def calculate_cost(current_cuts):
        total_cost = 0
        previous_cut = 0
        for cut in current_cuts:
            total_cost += cut - previous_cut
            previous_cut = cut
        
        total_cost += cake_length - previous_cut
        return total_cost

    def find_minimum_cost(index, current_cuts):
        # Base case: all cuts have been made
        if len(current_cuts) == number_of_cuts:
            return calculate_cost(current_cuts)
        
        minimum_cost = float('inf')
        
        # Explore all possible cut positions
        for cut_position in range(index, cake_length + 1):
            if cut_position not in current_cuts and cut_position in cuts:

                # Recursively explore the remaining cuts
                cost = find_minimum_cost(index + 1, current_cuts + [cut_position])
                minimum_cost = min(minimum_cost, cost)

        return minimum_cost

    # Start the recursion with no cuts made
    return find_minimum_cost(0, [])

Big(O) Analysis

Time Complexity

O(n!) – The described brute force approach considers all possible combinations of cuts. With 'n' being the number of required cuts, and given that each cut can be made at different positions on the cake, the first cut has roughly 'n' possibilities, the second cut roughly 'n-1' possibilities and so on. This leads to n * (n-1) * (n-2) ... * 1 possible combinations. This is equivalent to calculating n factorial, so the time complexity is O(n!).

Space Complexity

O(N) – The brute force approach explores all possible cut combinations through recursion. In the worst-case scenario, where the cuts are made in a highly unbalanced manner, the recursion depth can reach up to N, where N is the number of required cuts. Each level of recursion requires storing function call information (arguments, local variables, return address) on the call stack. Therefore, the auxiliary space used by the recursion stack scales linearly with the number of cuts, resulting in O(N) space complexity.

Optimal Solution

Approach

The problem involves finding the cheapest way to cut a cake given a set of cut locations. The key idea is to realize that the order in which you make the cuts affects the overall cost, and we can find the optimal order by thinking recursively.

Here's how the algorithm would work step-by-step:

First, imagine you've already made some cuts and are considering where to make the next one.
Realize that each cut divides a piece of cake into two smaller pieces.
The cost of a cut is equal to the length of the cake piece you are cutting.
To find the minimum cost, you need to explore all possible cut locations and figure out the cost for each of the cut location.
For each of the locations you considered, you now have 2 smaller pieces of cake and need to figure out the optimal way to cut each of them recursively.
Since you are using recursion, you will start by considering smaller pieces of cake and then working your way upwards.
Use a way to remember the smallest costs you find as you are exploring so you do not do the same calculation twice. This is a common strategy called 'memorization' that avoids repeating work, which is essential for making the process efficient.

Code Implementation

def min_cost_cut(cake_length, cuts):
    cuts = [0] + sorted(cuts) + [cake_length]
    number_of_cuts = len(cuts)

    memo = {}

    def calculate_cost(left_index, right_index):
        if right_index - left_index <= 1:
            return 0

        if (left_index, right_index) in memo:
            return memo[(left_index, right_index)]

        minimum_cost = float('inf')

        # Iterate through each possible cut location
        for i in range(left_index + 1, right_index):

            # Cost equals length of the current piece
            cost = cuts[right_index] - cuts[left_index]

            # Recursively calculate the cost of cutting the two pieces
            cost += calculate_cost(left_index, i) + calculate_cost(i, right_index)
            minimum_cost = min(minimum_cost, cost)

        # Store the calculated minimum cost
        memo[(left_index, right_index)] = minimum_cost

        return minimum_cost

    # Start the recursive calculation
    return calculate_cost(0, number_of_cuts - 1)

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm uses recursion with memoization to explore all possible cut orders. The core computation involves iterating through all possible cut locations within a given cake segment. With n cuts, there are at most n possible cut locations to consider for each cake segment. The memoization table stores results for all possible cake segments, which are defined by their start and end cut indices. Since there are n cuts, there are O(n^2) possible cake segments. For each segment, we iterate through O(n) cut locations. Therefore, the time complexity is O(n^3).

Space Complexity

O(N^2) – The dominant space usage comes from memoization. We are storing the minimum costs for cutting cake pieces of different sizes. The number of possible cake pieces is determined by all possible pairs of cut locations, which can be at most N^2 where N is the number of cut locations plus 2 boundary points. Therefore, the space complexity is O(N^2) for storing these intermediate results. The recursion stack space is O(N) but that is less than the memoization space.

Edge Cases

Case	How to Handle
Empty horizontal cuts or vertical cuts array	Return 0 as there are no cuts to be made and thus no cost.
Single horizontal or vertical cut	The algorithm should correctly calculate the cost based on the cake size and the single cut location.
Horizontal and vertical cuts at the same location (potentially multiple)	The cost calculation should correctly account for potentially overlapping regions.
Large number of cuts resulting in integer overflow in cost calculation	Use a data type with a larger range (e.g., long long in C++, long in Java) to prevent integer overflow.
Cuts that are out of bounds (negative or larger than cake dimensions)	Filter out invalid cuts before processing to avoid incorrect calculations.
Cake of size 1x1	Handle the case where width and height are one correctly, with a cost of 0 when there are no horizontal or vertical cuts.
Duplicate cut values within the horizontal or vertical cuts array	Handle duplicates correctly, as they represent cuts at the same location and should not be double-counted in the cost calculation.
Cuts provided already sorted	Ensure the algorithm still functions correctly even if horizontalCuts and verticalCuts are already sorted.

Empty horizontal cuts or vertical cuts array

How to Handle:

Return 0 as there are no cuts to be made and thus no cost.

Single horizontal or vertical cut

How to Handle:

The algorithm should correctly calculate the cost based on the cake size and the single cut location.

Horizontal and vertical cuts at the same location (potentially multiple)

How to Handle:

The cost calculation should correctly account for potentially overlapping regions.

Large number of cuts resulting in integer overflow in cost calculation

How to Handle:

Use a data type with a larger range (e.g., long long in C++, long in Java) to prevent integer overflow.

Cuts that are out of bounds (negative or larger than cake dimensions)

How to Handle:

Filter out invalid cuts before processing to avoid incorrect calculations.

Cake of size 1x1

How to Handle:

Handle the case where width and height are one correctly, with a cost of 0 when there are no horizontal or vertical cuts.

Duplicate cut values within the horizontal or vertical cuts array

How to Handle:

Handle duplicates correctly, as they represent cuts at the same location and should not be double-counted in the cost calculation.

Cuts provided already sorted

How to Handle:

Ensure the algorithm still functions correctly even if horizontalCuts and verticalCuts are already sorted.