Maximum Pizza Slices

Problem Understanding

We are given a circular array slices representing the sizes of pizza slices. We need to pick slices such that no two adjacent slices are picked (due to Alice and Bob picking adjacent slices). The goal is to maximize the total size of the slices we pick.

Naive Approach (Brute Force)

A brute-force approach would involve trying all possible combinations of picking slices such that no two adjacent slices are picked. However, this approach is highly inefficient due to the exponential number of combinations. It would involve recursion and backtracking, leading to a time complexity of roughly O(2^n), where n is the number of slices.

Due to its time complexity, this approach is not suitable for the given constraints.

Optimal Approach (Dynamic Programming)

We can solve this problem using dynamic programming. Since the pizza is circular, we can consider two cases:

1. The first slice is picked.
2. The first slice is not picked.

For each case, we can use dynamic programming to find the maximum sum of slice sizes we can pick without picking adjacent slices.

Let dp[i][j] be the maximum sum we can get by picking j slices from the first i slices.

The recurrence relation is as follows:

dp[i][j] = max(dp[i-1][j], slices[i] + dp[i-2][j-1])

where dp[i-1][j] represents not picking the ith slice, and slices[i] + dp[i-2][j-1] represents picking the ith slice. If we pick ith slice, we cannot pick i-1th slice, so we look back to i-2th slice and pick j-1 slices.

Algorithm:

1. Define a function maxSlices(slices) that takes the array of slice sizes as input.
2. Handle the base cases where the slices is null or its length is zero.
3. Calculate n, the length of slices array.
4. Calculate k = n/3, the number of slices we can take.
5. Create a new array without the last element to calculate the case without the first element being picked.
6. Create a new array without the first element to calculate the case with the first element being picked.
7. Calculate the maximum slice values with these two array using the DP approach.
8. Return the maximum value from these two cases.

Edge Cases:

• Empty input array: Return 0.
• Single element: Return the element.
• Array with two elements: Return the larger element

Code:

def maxSlices(slices):
    n = len(slices)
    k = n // 3

    def solve(arr):
        dp = [[0] * (k + 1) for _ in range(len(arr) + 1)]

        for i in range(1, len(arr) + 1):
            for j in range(1, k + 1):
                dp[i][j] = max(dp[i-1][j], arr[i-1] + (dp[i-2][j-1] if i > 1 else 0))
        return dp[len(arr)][k]

    case1 = solve(slices[:-1])
    case2 = solve(slices[1:])

    return max(case1, case2)

Time and Space Complexity:

• Time Complexity: O(n*k) where n is the number of slices and k = n/3 is the number of slices we are allowed to take. This is due to the dynamic programming table.
• Space Complexity: O(n*k) for the dynamic programming table.