human
I found an interesting problem on https://leetcode.com/problems/longest-turbulent-subarray/:
A subarray A[i], A[i+1], ..., A[j] of A is said to be turbulent if and only if:

For i <= k < j, A[k] > A[k+1] when k is odd, and A[k] < A[k+1] when k is even;
OR, for i <= k < j, A[k] > A[k+1] when k is even, and A[k] < A[k+1] when k is odd.

That is, the subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.
Return the length of a maximum size turbulent subarray of A.
 

Example 1:
Input: [9,4,2,10,7,8,8,1,9]
Output: 5
Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])

Example 2:
Input: [4,8,12,16]
Output: 2

Example 3:
Input: [100]
Output: 1



 
Note:

1 <= A.length <= 40000
0 <= A[i] <= 10^9

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
    def maxTurbulenceSize(self, A: List[int]) -> int:
        if len(A) < 2:
            return len(A)
        if len(A) == 2:
            return 2 if A[0] != A[1] else 1
        
        DP = [0 for _ in A]
        for i in range(1, len(A)):
            if DP[i-1] > 1:
                if A[i] > A[i-1]:
                    DP[i] = -DP[i-1] - 1
                elif A[i] < A[i-1]:
                    DP[i] = 2
                else:
                    DP[i] = 1
            elif DP[i-1] < -1:
                if A[i] < A[i-1]:
                    DP[i] = -DP[i-1] + 1
                elif A[i] > A[i-1]:
                    DP[i] = -2
                else:
                    DP[i] = -1
            else:
                if A[i] == A[i-1]:
                    DP[i] = 1
                else:
                    DP[i] = 2*(-1 if A[i] > A[i-1] else 1)
            
        return abs(max(DP, key = abs))