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LeetCode 1388: Pizza With 3n Slices Solution

Master LeetCode problem 1388 (Pizza With 3n Slices), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming with some constraints. We can consider the input array as a circular array and then use dynamic programming to keep track of the maximum sum we can obtain while making sure we don't select adjacent slices. We will break down the problem into two cases:

  1. If we consider the first slice.
  2. If we don't consider the first slice.

For each of the above cases, we can use dynamic programming to compute the maximum sum we can obtain and choose the maximum of the two. Finally, we return the maximum sum.

Algorithm:

  1. Create two arrays dp1 and dp2 of size n x n to store the maximum sum.
  2. Initialize dp1 and dp2 arrays based on the above two cases.
  3. Use dynamic programming to compute the maximum sum for each case.
  4. Return the maximum sum obtained.

Time Complexity:

The time complexity of this approach is O(n^2) where n is the number of slices.

Space Complexity:

The space complexity of this approach is O(n^2) for storing the dynamic programming arrays.

: :

Solution Code

class Solution {
    public int maxSizeSlices(int[] slices) {
        int n = slices.length;
        int k = n / 3;
        return Math.max(maxSum(slices, 0, n - 2 * k), maxSum(slices, 1, n - 2 * k + 1));
    }
    
    private int maxSum(int[] slices, int start, int end) {
        int n = slices.length;
        int k = (n + 1) / 3;
        int[][] dp = new int[k + 1][n / 3 + 1];
        
        for (int i = start; i < end; i++) {
            for (int j = k; j > 0; j--) {
                for (int m = 1; m <= n / 3; m++) {
                    dp[j][m] = Math.max(dp[j][m], dp[j - 1][m - 1] + slices[i]);
                }
            }
        }
        
        return dp[k][n / 3];
    }
}

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