1444 Number of Ways of Cutting a Pizza

🔀 Pick One

Given a rectangular pizza represented as a rows x cols matrix containing the following characters: 'A' (an apple) and '.' (empty cell) and given the integer k. You have to cut the pizza into k pieces using k-1 cuts.

For each cut you choose the direction: vertical or horizontal, then you choose a cut position at the cell boundary and cut the pizza into two pieces. If you cut the pizza vertically, give the left part of the pizza to a person. If you cut the pizza horizontally, give the upper part of the pizza to a person. Give the last piece of pizza to the last person.

Return the number of ways of cutting the pizza such that each piece contains at least one apple. Since the answer can be a huge number, return this modulo 10^9 + 7.

Example 1: 

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

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

Constraints:

  • 1 <= rows, cols <= 50
  • rows == pizza.length
  • cols == pizza[i].length
  • 1 <= k <= 10
  • pizza consists of characters 'A' and '.' only.
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        m, n = len(pizza), len(pizza[0])
        MOD = 10 ** 9 + 7
        # num of apples in pizza[i:][j:]
        pre_sum = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(m - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                pre_sum[i][j] = (
                    (1 if pizza[i][j] == 'A' else 0)
                    + pre_sum[i][j+1]
                    + pre_sum[i+1][j]
                    - pre_sum[i+1][j+1]
                )

        memo = {}

        def rec(i, j, cut):
            # rest of pizza: pizza[i:][j:]
            if pre_sum[i][j] == 0:
                # no apple left, 0 way to cut
                return 0
            if cut == 0:
                return 1
            if (i, j, cut) in memo:
                return memo[(i, j, cut)]
            res = 0
            # horizontal, possible to cut at [i+1, m-1]
            for ci in range(i + 1, m):
                # cut at ci
                if pre_sum[i][j] - pre_sum[ci][j] > 0:
                    res += rec(ci, j, cut-1)
                    res %= MOD
            # vertical, possible to cut [j+1, n-1]
            for cj in range(j + 1, n):
                # cut at cj
                if pre_sum[i][j] - pre_sum[i][cj] > 0:
                    res += rec(i, cj, cut-1)
                    res %= MOD
            memo[(i, j, cut)] = res
            return res
        return rec(0, 0, k - 1)