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Number of Ways of Cutting a Pizza

Difficulty: Hard

Problem Description

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.

Key Insights

  • The problem requires a way to partition a matrix while ensuring each partition contains at least one apple.
  • Dynamic programming (DP) is suitable for this problem due to its overlapping subproblems and optimal substructure properties.
  • Prefix sums can be used to efficiently check if a piece of pizza contains at least one apple.

Space and Time Complexity

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

Solution

To solve the problem, we use a dynamic programming approach where we maintain a DP table that keeps track of the number of ways to cut the pizza. We define dp[i][j][k] as the number of ways to cut the pizza starting from the cell (i, j) into k pieces.

  1. Prefix Sum Calculation: We first calculate a prefix sum to efficiently determine if a submatrix contains at least one apple.
  2. DP Initialization: We initialize our DP table and set up base cases for when k equals 1, where the entire pizza must contain at least one apple.
  3. Recursive DP Transition: For each possible cut (horizontal and vertical), we update our DP table based on previously computed values.

By iterating through possible cuts and using the prefix sum, we can efficiently compute the total number of valid ways to cut the pizza.

Code Solutions

def waysToCutPizza(pizza, k):
    rows, cols = len(pizza), len(pizza[0])
    MOD = 10**9 + 7
    
    # Create a prefix sum array
    prefix = [[0] * (cols + 1) for _ in range(rows + 1)]
    
    for i in range(rows - 1, -1, -1):
        for j in range(cols - 1, -1, -1):
            prefix[i][j] = (1 if pizza[i][j] == 'A' else 0) + prefix[i + 1][j] + prefix[i][j + 1] - prefix[i + 1][j + 1]

    # DP array
    dp = [[[0] * (k + 1) for _ in range(cols)] for _ in range(rows)]
    
    # Initialize base case when k = 1
    for i in range(rows):
        for j in range(cols):
            if prefix[i][j] > 0:  # there is at least one apple
                dp[i][j][1] = 1
    
    # Fill the DP table
    for cuts in range(2, k + 1):  # from 2 to k
        for i in range(rows):
            for j in range(cols):
                # Horizontal cuts
                for cut_row in range(i + 1, rows):
                    if prefix[i][j] - prefix[cut_row][j] > 0:  # If the upper part has apples
                        dp[i][j][cuts] = (dp[i][j][cuts] + dp[cut_row][j][cuts - 1]) % MOD
                
                # Vertical cuts
                for cut_col in range(j + 1, cols):
                    if prefix[i][j] - prefix[i][cut_col] > 0:  # If the left part has apples
                        dp[i][j][cuts] = (dp[i][j][cuts] + dp[i][cut_col][cuts - 1]) % MOD
    
    return dp[0][0][k]
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