Problem Description

You have an m x n rectangular cake that needs to be completely cut into 1 x 1 pieces.

The cake has m - 1 possible horizontal cut lines and n - 1 possible vertical cut lines. Each cut line has an associated cost:

• horizontalCut[i] represents the cost to make a cut along horizontal line i (where i ranges from 0 to m-2)
• verticalCut[j] represents the cost to make a cut along vertical line j (where j ranges from 0 to n-2)

When you make a cut:

1. You can choose any piece of cake that is not yet a 1 x 1 square
2. You can cut it either horizontally or vertically
3. The cut divides that piece into two separate pieces
4. The cost of making a cut along a specific line is always the same, regardless of how many times that line has been used in other pieces

The key insight is that when you cut along a line, you're actually cutting through all the pieces that the line passes through. For example:

• If you make a horizontal cut when the cake has already been divided into 3 vertical strips, you're cutting through all 3 strips, so the total cost is horizontalCut[i] * 3
• Similarly, if you make a vertical cut when the cake has 2 horizontal layers, the cost is verticalCut[j] * 2

Your goal is to find the minimum total cost to completely cut the cake into 1 x 1 pieces.

The solution uses a greedy approach: always make the most expensive cuts first (when they affect fewer pieces), which minimizes the total cost since expensive cuts will be multiplied by smaller numbers of pieces.

Intuition

Let's think about what happens when we make cuts. When we cut a horizontal line through the cake, we're actually cutting through all the vertical pieces that exist at that moment. If the cake has been divided into v vertical strips, then making a horizontal cut costs horizontalCut[i] * v. Similarly, a vertical cut through h horizontal layers costs verticalCut[j] * h.

The key observation is that each cut we make increases the number of pieces for future cuts. If we make a horizontal cut, we increase the number of horizontal layers by 1, which means all future vertical cuts will be more expensive. The same applies in reverse.

This creates a multiplication effect: if we have a cut with cost 10 and another with cost 5, the order matters significantly:

• Cut 10 first, then 5: 10 * 1 + 5 * 2 = 20
• Cut 5 first, then 10: 5 * 1 + 10 * 2 = 25

The more expensive cut should be made earlier when it affects fewer pieces!

This leads us to the greedy strategy: always make the most expensive cut available. By sorting both arrays in descending order and using two pointers, we can always choose the currently most expensive cut between horizontal and vertical options. Each time we make a cut, we update the count of how many pieces the perpendicular cuts will need to go through.

The beauty of this approach is that it guarantees the minimum cost because we're minimizing the multiplication factor for expensive cuts - they're always made when the number of pieces they need to cut through is as small as possible.

Learn more about Greedy, Dynamic Programming and Sorting patterns.

Solution Approach

The implementation follows a greedy algorithm with two pointers to efficiently select the most expensive cuts first.

Step 1: Sort the cut arrays

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

We sort both horizontalCut and verticalCut arrays in descending order so that the most expensive cuts appear first.

Step 2: Initialize variables

• ans = 0: Tracks the total cost
• i = j = 0: Pointers for traversing horizontalCut and verticalCut arrays
• h = v = 1: Initially, we have 1 horizontal piece and 1 vertical piece (the whole cake)

Step 3: Process cuts using two pointers

while i < m - 1 or j < n - 1:

We continue until all cuts are made (both arrays are fully traversed).

At each iteration, we decide whether to make a horizontal or vertical cut by comparing:

• If we've exhausted all vertical cuts (j == n - 1), we must make a horizontal cut
• If we still have both types available and horizontalCut[i] > verticalCut[j], we choose the horizontal cut (higher cost)
• Otherwise, we make a vertical cut

Step 4: Apply the selected cut

When making a horizontal cut:

ans += horizontalCut[i] * v
h, i = h + 1, i + 1

• The cost is horizontalCut[i] * v because this horizontal line cuts through v vertical pieces
• We increment h (number of horizontal layers increases by 1)
• We move to the next horizontal cut with i + 1

When making a vertical cut:

ans += verticalCut[j] * h
v, j = v + 1, j + 1

• The cost is verticalCut[j] * h because this vertical line cuts through h horizontal pieces
• We increment v (number of vertical strips increases by 1)
• We move to the next vertical cut with j + 1

Time Complexity: O(m log m + n log n) for sorting, then O(m + n) for the two-pointer traversal.

Space Complexity: O(1) if we don't count the sorting space, otherwise O(log m + log n) for the sorting algorithm.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: We have a 3×4 cake (m=3, n=4)

• horizontalCut = [1, 3] (costs for 2 horizontal cuts)
• verticalCut = [5, 2, 4] (costs for 3 vertical cuts)

Step 1: Sort arrays in descending order

• horizontalCut = [3, 1]
• verticalCut = [5, 4, 2]

Step 2: Initialize variables

• ans = 0 (total cost)
• i = 0, j = 0 (pointers)
• h = 1, v = 1 (initially one whole piece)

Step 3: Process cuts greedily

Iteration 1:

• Compare: verticalCut[0] = 5 vs horizontalCut[0] = 3
• Choose vertical cut (5 > 3)
• Cost: 5 × 1 = 5 (cutting through 1 horizontal piece)
• Update: ans = 5, v = 2, j = 1
• Cake is now in 2 vertical strips

Iteration 2:

• Compare: verticalCut[1] = 4 vs horizontalCut[0] = 3
• Choose vertical cut (4 > 3)
• Cost: 4 × 1 = 4 (cutting through 1 horizontal piece)
• Update: ans = 9, v = 3, j = 2
• Cake is now in 3 vertical strips

Iteration 3:

• Compare: verticalCut[2] = 2 vs horizontalCut[0] = 3
• Choose horizontal cut (3 > 2)
• Cost: 3 × 3 = 9 (cutting through 3 vertical strips!)
• Update: ans = 18, h = 2, i = 1
• Cake is now in 2 horizontal × 3 vertical = 6 pieces

Iteration 4:

• Compare: verticalCut[2] = 2 vs horizontalCut[1] = 1
• Choose vertical cut (2 > 1)
• Cost: 2 × 2 = 4 (cutting through 2 horizontal pieces)
• Update: ans = 22, v = 4, j = 3
• Cake is now in 2 horizontal × 4 vertical = 8 pieces

Iteration 5:

• Only horizontal cuts remain
• Choose horizontal cut
• Cost: 1 × 4 = 4 (cutting through 4 vertical strips)
• Update: ans = 26, h = 3, i = 2
• Cake is now in 3 horizontal × 4 vertical = 12 pieces (all 1×1)

Final Answer: 26

Notice how making expensive cuts early (when they affect fewer pieces) minimizes the total cost. If we had made the cheap cuts first, the expensive cuts would have been multiplied by larger numbers, resulting in a higher total cost.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × log m + n × log n)

The time complexity is dominated by the sorting operations:

• horizontalCut.sort(reverse=True) takes O((m-1) × log(m-1)) time since horizontalCut has m-1 elements
• verticalCut.sort(reverse=True) takes O((n-1) × log(n-1)) time since verticalCut has n-1 elements
• The while loop runs for exactly (m-1) + (n-1) iterations, performing O(1) operations in each iteration, contributing O(m + n) time

Since O((m-1) × log(m-1)) = O(m × log m) and O((n-1) × log(n-1)) = O(n × log n), and the sorting dominates the linear traversal, the overall time complexity is O(m × log m + n × log n).

Space Complexity: O(log m + log n)

The space complexity comes from:

• The sorting algorithm (typically Timsort in Python) uses O(log(m-1)) space for sorting horizontalCut and O(log(n-1)) space for sorting verticalCut in the worst case for the recursion stack
• The variables ans, i, j, h, and v use O(1) space

Therefore, the overall space complexity is O(log m + log n).

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

Common Pitfalls

1. Misunderstanding the Cut Multiplication Logic

A frequent mistake is thinking that each cut has a fixed cost regardless of when it's made. In reality, the cost of a cut depends on how many pieces it passes through at the time of cutting.

Incorrect Understanding:

# WRONG: Simply summing all cut costs
total_cost = sum(horizontalCut) + sum(verticalCut)

Correct Understanding:

• When you make a horizontal cut after k vertical cuts have been made, the horizontal cut passes through k+1 vertical pieces
• The actual cost is horizontalCut[i] * (number_of_vertical_pieces)

2. Incorrect Greedy Choice - Choosing Cheapest Cuts First

Some might intuitively think to minimize cost by making cheaper cuts first, but this is backwards.

Wrong Approach:

# WRONG: Sorting in ascending order (cheapest first)
horizontalCut.sort()  # Should be reverse=True
verticalCut.sort()    # Should be reverse=True

Why it's wrong: If you make cheap cuts first, expensive cuts will be made later when there are more pieces, multiplying their cost by larger numbers.

Correct Strategy: Make expensive cuts first when there are fewer pieces to minimize their multiplication factor.

3. Off-by-One Errors in Piece Counting

A subtle but critical error is incorrectly tracking the number of pieces.

Common Mistake:

# WRONG: Starting with 0 pieces
h_pieces = 0  # Should start with 1
v_pieces = 0  # Should start with 1

Why it matters:

• Initially, you have 1 whole cake (not 0 pieces)
• After 1 cut, you have 2 pieces (not 1)
• After k cuts in one direction, you have k+1 pieces in that direction

4. Index Boundary Confusion

Mixing up the relationship between cake dimensions and number of cuts.

Wrong:

# WRONG: Using m and n as the number of cuts
while h_index < m or v_index < n:  # Should be m-1 and n-1

Remember:

• An m x n cake requires m-1 horizontal cuts and n-1 vertical cuts
• The arrays horizontalCut and verticalCut have lengths m-1 and n-1 respectively

5. Incorrect Comparison Logic in the Greedy Decision

The condition for choosing which cut to make next must handle edge cases properly.

Incomplete Logic:

# WRONG: Doesn't handle when one array is exhausted
if horizontalCut[h_index] > verticalCut[v_index]:
    # Make horizontal cut

Correct Logic:

if v_index == n - 1 or (h_index < m - 1 and horizontalCut[h_index] > verticalCut[v_index]):
    # Make horizontal cut

The condition must check:

1. If vertical cuts are exhausted (v_index == n - 1), must make horizontal cut
2. If both cuts are available, choose the more expensive one
3. Otherwise, make the vertical cut (including when horizontal cuts are exhausted)