How do you solve Number of Ways of Cutting a Pizza on LeetCode?

Number of Ways of Cutting a Pizza (LeetCode #1444) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² * k) time complexity and O(n²) space complexity. Key insight: Understanding prefix sums helps in quickly checking for apples in subregions.

What is the brute force solution for Number of Ways of Cutting a Pizza?

The brute force approach for Number of Ways of Cutting a Pizza is Brute Force, with O(n² * k) time complexity and O(1) space complexity. The brute force approach involves trying all possible ways to cut the pizza into k pieces. We can recursively explore each possible cut and check if each resulting piece contains at least one apple. The optimal Optimal Solution approach improves this to O(n² * k).

What algorithm or data structure is used to solve Number of Ways of Cutting a Pizza?

Number of Ways of Cutting a Pizza uses the following concepts: Array, Dynamic Programming, Memoization, Matrix, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Memoization, Prefix Sum.

What are the interview tips for Number of Ways of Cutting a Pizza?

Explain your thought process clearly while coding. Discuss edge cases, such as when k = 1 or when there are no apples. Practice writing clean, modular code to improve readability.

What are common mistakes when solving Number of Ways of Cutting a Pizza?

Not using memoization leading to excessive recursion and timeouts. Failing to check if a piece contains apples before counting it.

#1444

Number of Ways of Cutting a Pizza

Hard

ArrayDynamic ProgrammingMemoizationMatrixPrefix SumDynamic ProgrammingMemoizationPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² * k)	O(n² * k)
Space	O(1)	O(n²)

💡

Intuition

Time O(n² * k)Space O(n²)

The optimal approach uses dynamic programming to store results of subproblems, reducing redundant calculations. We maintain a prefix sum array to quickly check if a piece contains apples.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum array to count apples in any subrectangle of the pizza.
2Step 2: Use a recursive function with memoization to count ways to cut the pizza, checking each possible cut.
3Step 3: Return the result from the memoized function.

solution.py34 lines

1# Full working Python code
2from typing import List
3
4class Solution:
5    def waysToCutPizza(self, pizza: List[str], k: int) -> int:
6        MOD = 10**9 + 7
7        rows, cols = len(pizza), len(pizza[0])
8        prefix = [[0] * (cols + 1) for _ in range(rows + 1)]
9
10        # Calculate prefix sum of apples
11        for r in range(rows - 1, -1, -1):
12            for c in range(cols - 1, -1, -1):
13                prefix[r][c] = (1 if pizza[r][c] == 'A' else 0) + prefix[r + 1][c] + prefix[r][c + 1] - prefix[r + 1][c + 1]
14
15        memo = {}  # Memoization dictionary
16
17        def dfs(r1, c1, cuts_left):
18            if cuts_left == 0:
19                return 1 if prefix[r1][c1] > 0 else 0
20            if (r1, c1, cuts_left) in memo:
21                return memo[(r1, c1, cuts_left)]
22            ways = 0
23            # Horizontal cuts
24            for r in range(r1 + 1, rows):
25                if prefix[r1][c1] - prefix[r][c1] > 0:
26                    ways = (ways + dfs(r, c1, cuts_left - 1)) % MOD
27            # Vertical cuts
28            for c in range(c1 + 1, cols):
29                if prefix[r1][c1] - prefix[r1][c] > 0:
30                    ways = (ways + dfs(r1, c, cuts_left - 1)) % MOD
31            memo[(r1, c1, cuts_left)] = ways
32            return ways
33
34        return dfs(0, 0, k - 1)

ℹ

Complexity note: The complexity arises from calculating the prefix sums and the recursive calls, but memoization significantly reduces redundant calculations.

1Understanding prefix sums helps in quickly checking for apples in subregions.
2Dynamic programming with memoization avoids recalculating results for the same state.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.