#1444

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Number of Ways of Cutting a Pizza

This problem challenges you to determine the number of valid ways to cut a pizza into pieces with apples using dynamic programming.

array dynamic programming memoization matrix prefix sum

Problem Statement

You are given a rectangular pizza represented by a matrix of 'A' (apple) and '.' (empty) cells. Your task is to divide the pizza into k pieces using k-1 cuts. Each cut is either vertical or horizontal, and the pieces must contain at least one apple. After each cut, the remaining pizza's bottom-right corner must be (rows-1, cols-1).

Return the number of ways to cut the pizza such that every piece contains at least one apple. Since the result can be large, return the number modulo 10^9 + 7. Consider using dynamic programming and memoization to optimize your approach.

Examples

Example 1

Input: pizza = ["A..","AAA","..."], k = 3

Output: 3

The figure above shows the three ways to cut the pizza. Note that pieces must contain at least one apple.

Example 2

Input: pizza = ["A..","AA.","..."], k = 3

Output: 1

Example details omitted.

Example 3

Input: pizza = ["A..","A..","..."], k = 1

Output: 1

Example details omitted.

Constraints

1 <= rows, cols <= 50
rows == pizza.length
cols == pizza[i].length
1 <= k <= 10
pizza consists of characters 'A' and '.' only.

Solution Approach

State Transition with Dynamic Programming

The key observation in this problem is to break down the pizza cuts into subproblems using dynamic programming. We track the state of each subproblem using the pizza matrix's positions, specifically the bottom-right corner. We can recursively explore cutting the pizza at every possible vertical or horizontal position while ensuring each segment has at least one apple.

Memoization for Optimization

Since the problem involves overlapping subproblems, memoization can be used to store previously computed results. For each state (pizza portion and remaining cuts), we can store the result in a memoization table to avoid redundant computations and speed up the solution.

Handling Multiple Cuts and Directions

We need to handle both horizontal and vertical cuts. For each cut, explore all valid ways of dividing the pizza by cutting at every possible position. After each cut, update the remaining pizza and recursively calculate the number of valid ways to divide the pizza until all cuts are made.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the number of possible subproblems (pizza pieces) and the number of cuts left to make. Given that there are up to 50 rows and 50 columns, and we may need to make up to 10 cuts, the problem complexity grows rapidly. However, memoization significantly reduces redundant computations, making the approach feasible.

What Interviewers Usually Probe

The candidate should demonstrate a clear understanding of dynamic programming principles, especially state transitions and recursion.
A strong candidate will use memoization to optimize their solution and avoid recomputing subproblems.
Look for a solution that correctly handles both horizontal and vertical cuts and considers the modular arithmetic to avoid overflow.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly handle overlapping subproblems, resulting in unnecessary recomputation and poor performance.
Not properly considering all possible positions for cuts, which may lead to incomplete solutions.
Overlooking the modular arithmetic for large results, which can lead to incorrect outputs due to overflow.

Follow-up variants

Try solving this problem with different constraints, such as increasing the number of cuts or the size of the pizza.
Consider alternative ways of dividing the pizza, such as allowing diagonal cuts or limiting cuts to only one direction.
Explore solutions that minimize the time complexity by using different state representations or iterative techniques.

FAQ

How do I approach the state transition for this pizza problem?

The key is to break down the problem into smaller subproblems, tracking the state of each piece of pizza after each cut and considering whether it contains an apple. Recursion and memoization are critical here.

What is the role of dynamic programming in solving this problem?

Dynamic programming is used to break down the problem into manageable subproblems. We store previously computed results to avoid redundant calculations, improving efficiency.

What is the modular arithmetic requirement in this problem?

Since the number of ways to cut the pizza can be large, we return the result modulo 10^9 + 7 to avoid overflow and ensure correctness.

How does memoization improve performance in this problem?

Memoization ensures that each subproblem is computed only once, storing the result for future reference. This significantly reduces the time complexity of the solution.

What happens if I forget to handle all possible cut positions?

Not considering all valid cut positions could lead to missing possible ways to divide the pizza, resulting in an incomplete solution.

terminal

Solution

Solution 1: 2D Prefix Sum + Memoized Search

We can use a 2D prefix sum to quickly calculate the number of apples in each sub-rectangle. Define $s[i][j]$ to represent the number of apples in the sub-rectangle that includes the first $i$ rows and the first $j$ columns. Then $s[i][j]$ can be derived from the number of apples in the three sub-rectangles $s[i-1][j]$, $s[i][j-1]$, and $s[i-1][j-1]$. The specific calculation method is as follows:

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        @cache
        def dfs(i: int, j: int, k: int) -> int:
            if k == 0:
                return int(s[m][n] - s[i][n] - s[m][j] + s[i][j] > 0)
            ans = 0
            for x in range(i + 1, m):
                if s[x][n] - s[i][n] - s[x][j] + s[i][j] > 0:
                    ans += dfs(x, j, k - 1)
            for y in range(j + 1, n):
                if s[m][y] - s[i][y] - s[m][j] + s[i][j] > 0:
                    ans += dfs(i, y, k - 1)
            return ans % mod

        mod = 10**9 + 7
        m, n = len(pizza), len(pizza[0])
        s = [[0] * (n + 1) for _ in range(m + 1)]
        for i, row in enumerate(pizza, 1):
            for j, c in enumerate(row, 1):
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + int(c == 'A')
        return dfs(0, 0, k - 1)

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