Introduction

In this blog post, we will explore how to solve the LeetCode problem "1444" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's a breakdown of the solution:

• Prefix Sum for Apple Counts: Create a 2D prefix sum array to efficiently determine the number of apples within any rectangular region of the pizza. This avoids repeatedly iterating through sub-matrices.
• Dynamic Programming: Use a 3D DP array dp[k][row][col] to store the number of ways to cut the pizza into k pieces starting from row row and column col.

• Recursive Calculation: Recursively explore possible horizontal and vertical cuts. Check if each cut results in a piece with at least one apple using the prefix sum array.

• Time Complexity: O(k * rows * cols * (rows + cols))

• Space Complexity: O(k * rows * cols)

def solve():
    def ways_to_cut_pizza(pizza, k):
        rows = len(pizza)
        cols = len(pizza[0])
        MOD = 10**9 + 7

        # apples[i][j] stores the number of apples in pizza[i:][j:]
        apples = [[0] * (cols + 1) for _ in range(rows + 1)]
        for i in range(rows - 1, -1, -1):
            for j in range(cols - 1, -1, -1):
                apples[i][j] = apples[i + 1][j] + apples[i][j + 1] - apples[i + 1][j + 1] + (1 if pizza[i][j] == 'A' else 0)

        # dp[k][i][j] stores the number of ways to cut the pizza into k pieces starting from pizza[i:][j:]
        dp = [[[0] * cols for _ in range(rows)] for _ in range(k)]

        # Base case: If we only need one piece, there is one way to cut it if it contains at least one apple
        for i in range(rows):
            for j in range(cols):
                if apples[i][j] > 0:
                    dp[0][i][j] = 1

        # Iterate over the number of pieces from 2 to k
        for remain in range(1, k):
            for i in range(rows):
                for j in range(cols):
                    # Horizontal cuts
                    for next_i in range(i + 1, rows):
                        # Check if the upper part contains at least one apple
                        if apples[i][j] - apples[next_i][j] > 0:
                            dp[remain][i][j] = (dp[remain][i][j] + dp[remain - 1][next_i][j]) % MOD

                    # Vertical cuts
                    for next_j in range(j + 1, cols):
                        # Check if the left part contains at least one apple
                        if apples[i][j] - apples[i][next_j] > 0:
                            dp[remain][i][j] = (dp[remain][i][j] + dp[remain - 1][i][next_j]) % MOD

        return dp[k - 1][0][0]

    pizza = ["A..","AAA","..."]
    k = 3
    print(ways_to_cut_pizza(pizza, k))

    pizza = ["A..","AA.","..."]
    k = 3
    print(ways_to_cut_pizza(pizza, k))

    pizza = ["A..","A..","..."]
    k = 1
    print(ways_to_cut_pizza(pizza, k))


    # Code
    ```python
    def ways_to_ut_pizza(pizza, k):    rows = len(pizza)
    cols = len(pizza[0])
    MOD = 10**9 + 7

    # apples[i][j] stores the number of apples in pizza[i:][j:]
    apples = [[0] * (cols + 1) for _ in range(rows + 1)]
    for i in range(rows - 1, -1, -1):
        for j in range(cols - 1, -1, -1):
            apples[i][j] = apples[i + 1][j] + apples[i][j + 1] - apples[i + 1][j + 1] + (1 if pizza[i][j] == 'A' else 0)

    # dp[k][i][j] stores the number of ways to cut the pizza into k pieces starting from pizza[i:][j:]
    dp = [[[0] * cols for _ in range(rows)] for _ in range(k)]

    # Base case: If we only need one piece, there is one way to cut it if it contains at least one apple
    for i in range(rows):
        for j in range(cols):
            if apples[i][j] > 0:
                dp[0][i][j] = 1

    # Iterate over the number of pieces from 2 to k
    for remain in range(1, k):
        for i in range(rows):
            for j in range(cols):
                # Horizontal cuts
                for next_i in range(i + 1, rows):
                    # Check if the upper part contains at least one apple
                    if apples[i][j] - apples[next_i][j] > 0:
                        dp[remain][i][j] = (dp[remain][i][j] + dp[remain - 1][next_i][j]) % MOD

                # Vertical cuts
                for next_j in range(j + 1, cols):
                    # Check if the left part contains at least one apple
                    if apples[i][j] - apples[i][next_j] > 0:
                        dp[remain][i][j] = (dp[remain][i][j] + dp[remain - 1][i][next_j]) % MOD

    return dp[k - 1][0][0]