human
Give me a the Python code you requested given the following:
I found an interesting problem on https://leetcode.com/problems/arithmetic-slices/:
A sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.

For example, these are arithmetic sequence:
1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9

The following sequence is not arithmetic. 1, 1, 2, 5, 7 

A zero-indexed array A consisting of N numbers is given. A slice of that array is any pair of integers (P, Q) such that 0 

A slice (P, Q) of array A is called arithmetic if the sequence:
    A[P], A[p + 1], ..., A[Q - 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.

The function should return the number of arithmetic slices in the array A. 

Example:

A = [1, 2, 3, 4]

return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.

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 numberOfArithmeticSlices(self, A):
         curr, sum = 0, 0
         for i in range(2,len(A)):
             if A[i]-A[i-1] == A[i-1]-A[i-2]:
                 curr += 1
                 sum += curr
             else:
                 curr = 0
         return sum
 #         solution = 0
 #         connected = 1
 #         old_diff = None
 #         sequences = []
 #         if len(A) < 3:
 #             return 0
         
 #         for index,num in enumerate(A):
 #             if index < len(A) - 1:
 #                 new_diff = num - A[index + 1]
 #             else:
 #                 new_diff = A[index - 1] - num
 #             if old_diff == new_diff:
 #                 if index == len(A) - 1 and connected >= 3:
 #                     connected += 1
 #                     sequences.append(connected)
 #                 connected += 1
 #             else:
 #                 old_diff = new_diff
 #                 if connected > 2:
 #                     sequences.append(connected)
 #                 connected = 1
 #         for sequence in sequences:
 #             prev = 0
 #             while sequence >= 2:
 #                 prev += 1
 #                 solution += prev
 #                 sequence -= 1
 #         return solution