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Number of Ways of Cutting a Pizza

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Dynamic Programming

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.

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the dimensions (number of rows and columns) of the pizza, and what are the constraints on these dimensions?
  2. What characters can appear in the pizza string array, and are there any restrictions on their frequency (e.g., is it guaranteed there will always be at least one 'A'?)?
  3. What is the range for 'k', the number of cuts allowed?
  4. If it's impossible to make 'k' cuts such that each piece has at least one apple, what value should I return?
  5. Does the order of the cuts matter? In other words, is cutting horizontally then vertically the same as cutting vertically then horizontally?

Brute Force Solution

Approach

The brute force approach to cutting a pizza is all about trying every possible combination of cuts. We explore all possible vertical and horizontal cuts, checking if a valid cut arrangement can be achieved. This involves checking whether each piece has at least one apple.

Here's how the algorithm would work step-by-step:

  1. Imagine you have the whole pizza in front of you, and you need to make a certain number of cuts.
  2. Start by trying to make the first cut in every possible location either vertically or horizontally.
  3. For each of these first cuts, consider every possible place for the second cut, both vertically and horizontally.
  4. Keep doing this, trying all possible locations for each cut, one after the other, until you've made the required number of cuts.
  5. After making all the cuts in a particular arrangement, check each piece of the pizza to see if it contains at least one apple.
  6. If every piece has at least one apple, then this is a valid way of cutting the pizza; otherwise, it is not valid.
  7. Keep track of every valid way you find to cut the pizza.
  8. Once you have considered all possible arrangements of cuts, count the total number of valid ways you found.

Code Implementation

def number_of_ways_of_cutting_a_pizza_brute_force(pizza, number_of_cuts):
    rows = len(pizza)
    columns = len(pizza[0])
    number_of_valid_ways = 0

    def is_valid_cut(pieces):
        for piece in pieces:
            has_apple = False
            for row_index in range(len(piece)):
                for column_index in range(len(piece[0])):
                    if piece[row_index][column_index] == 'A':
                        has_apple = True
                        break
                if has_apple:
                    break
            if not has_apple:
                return False
        return True

    def generate_cuts(current_cuts):
        nonlocal number_of_valid_ways

        if len(current_cuts) == number_of_cuts:
            # After all cuts are placed, divide the pizza into pieces
            pieces = divide_pizza(current_cuts)

            if is_valid_cut(pieces):
                number_of_valid_ways += 1
            return

        # Iterate over all possible row and column cut positions
        for row_cut in range(1, rows):
            generate_cuts(current_cuts + [('H', row_cut)])
        
        for column_cut in range(1, columns):
            generate_cuts(current_cuts + [('V', column_cut)])

    def divide_pizza(cuts):
        cuts.sort(key=lambda x: x[1])
        horizontal_cuts = [cut[1] for cut in cuts if cut[0] == 'H']
        vertical_cuts = [cut[1] for cut in cuts if cut[0] == 'V']

        pieces = []
        row_start = 0

        # Split the pizza by each horizontal cut
        for row_end in horizontal_cuts + [rows]:
            column_start = 0
            # Split the pizza by each vertical cut
            for column_end in vertical_cuts + [columns]:
                piece = []

                for row_index in range(row_start, row_end):
                    row_segment = pizza[row_index][column_start:column_end]
                    piece.append(row_segment)

                pieces.append(piece)
                column_start = column_end
            row_start = row_end

        return pieces

    generate_cuts([])
    return number_of_valid_ways

Big(O) Analysis

Time Complexity
O(m^k * n^k)The brute force approach iterates through all possible combinations of cuts. We have an m x n pizza and need to make k cuts. For each of the k cuts, we can either make a horizontal cut in one of the m-1 possible locations or a vertical cut in one of the n-1 possible locations. Therefore, for each cut, we have approximately m + n possibilities, which is further simplified to just 'n' since we are dealing with the general growth rate and consider m and n to be in the same order of magnitude. Because this is repeated k times, and for each combination of cuts we need to verify if each slice has at least one apple which takes O(m * n), the overall time complexity can be approximated as O((m*n)^k), but given we are mainly concerned with the cutting choices, this can be seen as O(m^k * n^k) since we make a choice among m horizontal and n vertical positions for each of k cuts. Verifying each slice also scales linearly, but is dominated by the cut choices.
Space Complexity
O(K)The brute force approach described uses recursion to try all possible cut combinations. The maximum depth of this recursion corresponds to the number of cuts, K, as each level explores a new cut. Each recursive call requires storing local variables and the call stack frame. Therefore, the auxiliary space required grows linearly with the number of cuts K, leading to a space complexity of O(K).

Optimal Solution

Approach

The most efficient approach to counting pizza cuts avoids recalculating the same information multiple times. We leverage a technique where we store and reuse previous computations to determine the number of valid cuts for each possible sub-pizza. This significantly speeds up the process.

Here's how the algorithm would work step-by-step:

  1. First, determine the crucial ingredient presence in the original pizza as well as all possible sub-pizza sections. This is pre-computed to avoid redundant checks later.
  2. Imagine working backward from your desired number of cuts. Start by understanding that the last cut must result in a final piece with at least one crucial ingredient.
  3. Recognize that the number of ways to make a cut is linked to the number of ways to make one fewer cut on the remaining larger pizza section.
  4. Store the results of calculating the number of valid cut combinations for each sub-pizza and number of cuts. This avoids redundant calculations.
  5. When considering a new cut, see if the result has already been calculated and stored. If so, reuse the stored result to save time.
  6. Continue this process for the number of cuts, building upon previously calculated results until you arrive at the first cut of the full pizza.
  7. The final result will be the number of ways to cut the pizza into the desired number of pieces, where each piece has at least one crucial ingredient.

Code Implementation

def number_of_ways_of_cutting_a_pizza(pizza, number_of_cuts_needed):
    rows = len(pizza)
    cols = len(pizza[0])
    modulo = 10**9 + 7

    prefix_sum = [[0] * (cols + 1) for _ in range(rows + 1)]
    for row_index in range(1, rows + 1):
        for col_index in range(1, cols + 1):
            prefix_sum[row_index][col_index] = prefix_sum[row_index - 1][col_index] + prefix_sum[row_index][col_index - 1] - prefix_sum[row_index - 1][col_index - 1] + (pizza[row_index - 1][col_index - 1] == 'A')

    memo = {}

    def has_ingredient(row_start, col_start, row_end, col_end):
        return prefix_sum[row_end + 1][col_end + 1] - prefix_sum[row_start][col_end + 1] - prefix_sum[row_end + 1][col_start] + prefix_sum[row_start][col_start] > 0

    def solve(remaining_cuts, row_start, col_start):
        # Base cases: no more cuts needed
        if remaining_cuts == 0:
            return 1 if has_ingredient(row_start, col_start, rows - 1, cols - 1) else 0

        # Check if the result is already memoized
        if (remaining_cuts, row_start, col_start) in memo:
            return memo[(remaining_cuts, row_start, col_start)]

        ways = 0

        # Horizontal cuts
        for row_index in range(row_start, rows - 1):
            # Ensuring each piece has at least one 'A'
            if has_ingredient(row_start, col_start, row_index, cols - 1):
                ways = (ways + solve(remaining_cuts - 1, row_index + 1, col_start)) % modulo

        # Vertical cuts
        for col_index in range(col_start, cols - 1):
            # Ensuring each piece has at least one 'A'
            if has_ingredient(row_start, col_start, rows - 1, col_index):
                ways = (ways + solve(remaining_cuts - 1, row_start, col_index + 1)) % modulo

        # Store the result in the memo
        memo[(remaining_cuts, row_start, col_start)] = ways
        return ways

    # We need to handle the initial state where 0 cuts are required, and there are no ingredients.
    if not has_ingredient(0, 0, rows - 1, cols - 1) and number_of_cuts_needed >= 0:
      return 0

    # Initiate the recursive process with the full pizza
    result = solve(number_of_cuts_needed, 0, 0)
    return result

Big(O) Analysis

Time Complexity
O(m * n * k)The core of the algorithm involves a 3D DP table (memoization) to store the number of ways to cut the pizza. The dimensions of this table are determined by the number of rows (m), the number of columns (n), and the number of cuts (k). We iterate through all possible sub-pizzas defined by their top-left corner and bottom-right corner implicitly during cut considerations. Calculating the prefix sum for crucial ingredient counts is O(m * n) upfront, but the recurrence relation in the DP table is visited m * n * k times. Therefore, the overall time complexity is dominated by filling the DP table, leading to O(m * n * k).
Space Complexity
O(m * n * k)The algorithm uses a 3D memoization table (dynamic programming table) to store the results of previously calculated subproblems. The dimensions of this table are determined by the number of rows (m) and columns (n) of the pizza, and the number of allowed cuts (k). Therefore, the auxiliary space required to store this table is proportional to m * n * k, where m is the number of rows, n is the number of columns, and k is the number of cuts. The prefix sum matrix to compute ingredient presence also takes O(m * n) space, but this is dominated by the memoization table's space complexity. Thus, the overall space complexity is O(m * n * k).

Edge Cases

Null or empty pizza array
How to Handle:
Return 0 since there's no pizza to cut.
k (number of cuts) is zero
How to Handle:
Return 1 if there is at least one apple, 0 otherwise.
k (number of cuts) is greater than the rows or columns -1
How to Handle:
Return 0 since it is impossible to make that many cuts.
Pizza with no apples
How to Handle:
Return 0 since no cuts will result in a valid pizza.
Large pizza dimensions exceeding memory constraints
How to Handle:
Optimize memory usage by using memoization effectively and considering data type sizes to avoid overflow.
Integer overflow when calculating intermediate sums or results
How to Handle:
Use modulo operator (%) during calculations to prevent integer overflow.
Only one row/column in the pizza array
How to Handle:
Handle separately since slicing can only occur in the larger dimension with available cuts.
All cells contain apples
How to Handle:
Calculate all possible combinations of cuts subject to valid cut counts for correct answer.