Problem Description

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:

1. Cut along a horizontal line i at a cost of horizontalCut[i].
2. 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.

Intuition

To solve the problem of minimizing the total cost to cut the cake into 1 x 1 pieces, we must understand that the arrangement and order of cuts determine the total cost. Specifically, making cuts with higher costs earlier can be advantageous, as they will apply to more pieces initially.

The key idea is as follows:

• Maximize the Impact of Expensive Cuts: Each cut divides the cake into additional smaller pieces. Thus, if we apply a costly cut early, it affects a larger number of resulting pieces for future cuts. The cost of making a horizontal cut is multiplied by the number of vertical sections (and vice versa). Therefore, making the highest cost cut at each step ensures minimizing total expense when multiplied by its impact.

To implement this strategy:

1. Sort the Cuts: First, arrange both horizontalCut and verticalCut in descending order. This allows us to use a greedy approach in selecting the most costly cut at each step.
2. Use Two Pointers: We maintain two pointers, one for horizontalCut and one for verticalCut. At every step, check which cut has a greater cost and apply it, adjusting the product terms (h or v—number of vertical sections when making a horizontal cut, or number of horizontal sections when making a vertical cut).
3. Increment and Update: Whenever a cut is made, update the respective number of pieces divided, and move to the next cut option by incrementing the pointers.

By following this approach, both horizontal and vertical cuts are handled optimally with respect to their cost and impact on the overall division of the cake. The solution involves sorting, decision-making based on maximum cost impact, and updates that reflect the current number of divisions. This effectively minimizes the total cost required to cut the entire cake into 1 x 1 pieces.

Learn more about Greedy, Dynamic Programming and Sorting patterns.

Solution Approach

The solution to minimize the cost of cutting the entire cake into 1 x 1 pieces involves using a greedy algorithm combined with sorting and two-pointer techniques. Here's how the approach is implemented step-by-step:

1. Sorting:

   • Start by sorting both horizontalCut and verticalCut arrays in descending order. This prepares us to consider the most costly cuts first, which is crucial for minimizing the total cost.

   horizontalCut.sort(reverse=True)
   verticalCut.sort(reverse=True)

2. Initialization:

   • Initialize variables to keep track of the total cost (ans), pointers for the horizontal and vertical cut indices (i and j), and the number of horizontal and vertical parts (h and v). Initially, the entire cake is one whole piece, hence h = 1 and v = 1.

   ans = i = j = 0
   h = v = 1

3. Two Pointers and Greedy Selection:

   • Use a while loop to iterate through until all possible cuts are considered. In each iteration:
     • Compare the current highest horizontal and vertical cut costs.
     • Make the cut with the higher cost to minimize future expenses. Adjust the respective horizontal or vertical part count.
     • If a horizontal cut is made, update the equation ans += horizontalCut[i] * v.
     • If a vertical cut is made, update the equation ans += verticalCut[j] * h.
     • Increment the appropriate pointer (i for horizontal cuts and j for vertical cuts) and the respective segment count (h or v).

   while i < m - 1 or j < n - 1:
       if j == n - 1 or (i < m - 1 and horizontalCut[i] > verticalCut[j]):
           ans += horizontalCut[i] * v
           h, i = h + 1, i + 1
       else:
           ans += verticalCut[j] * h
           v, j = v + 1, j + 1

4. Return the Total Minimum Cost:

   • At the end of the loop, ans contains the minimum total cost for cutting the cake into 1 x 1 pieces.

   return ans

This approach effectively uses the properties of greedy algorithms by minimizing cost at each step through strategic selection of cuts, ensuring that the most expensive cut impacts the largest current piece size. The use of sorting, two-pointer technique, and greedy selection ensures an optimal solution.

Example Walkthrough

Let's illustrate the solution approach with a small example:

Consider a cake of size 3 x 3 (so m = 3 and n = 3). We are given two arrays representing the costs of cuts:

• horizontalCut = [2, 3]
• verticalCut = [1, 4]

Our goal is to cut the cake into 1 x 1 pieces while minimizing the total cost.

1. Sorting:

We start by sorting both horizontalCut and verticalCut in descending order:

• horizontalCut becomes [3, 2]
• verticalCut becomes [4, 1]

2. Initialization:

We initialize the total cost ans to 0, and set the indices for the horizontal and vertical cuts i and j to 0. We also initialize the number of horizontal sections h and vertical sections v to 1.

ans = i = j = 0
h = v = 1

3. Two Pointers and Greedy Selection:

We process the cuts using a while loop, selecting the more costly option at each iteration:

• Iteration 1:

  • horizontalCut[0] is 3, verticalCut[0] is 4.
  • Choose the vertical cut since 4 > 3.
  • Update: ans += 4 * 1 = 4, increment v to 2, move to the next vertical cut by setting j = 1.

• Iteration 2:

  • horizontalCut[0] is 3, verticalCut[1] is 1.
  • Choose the horizontal cut since 3 > 1.
  • Update: ans += 3 * 2 = 6, increment h to 2, move to the next horizontal cut by setting i = 1.

• Iteration 3:

  • horizontalCut[1] is 2, verticalCut[1] is 1.
  • Choose the horizontal cut since 2 > 1.
  • Update: ans += 2 * 2 = 4, increment h to 3, i reaches 2 which is the limit for horizontal cuts.

• Iteration 4:

  • Now, all horizontal cuts are done. Focus on the remaining vertical cut.
  • Choose the vertical cut verticalCut[1], which is 1.
  • Update: ans += 1 * 3 = 3, increment v to 3, j reaches 2 which is the limit for vertical cuts.

Conclusion:

Now, the entire cake is divided into 1 x 1 pieces. The total minimum cost ans is 17.

return ans  # Final result is 17

By following this example, we can see how each decision at each step impacts the total cost based on the current number of sections affected by the cuts.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(m * log m + n * log n). This is due to the sorting operations on the horizontalCut and verticalCut lists, which each require O(m * log m) and O(n * log n) time respectively. The subsequent while-loop contributes O(m + n) complexity since at most m + n iterations are needed, but this is dominated by the sorting term.

The space complexity is O(log m + log n), which is attributed to the auxiliary space needed for sorting the two lists. Each sort operation can require up to O(log m) and O(log n) space, cumulatively.

Learn more about how to find time and space complexity quickly.