human
Give me a the Python code you requested given the following:
Solve in Python:
Given an array A, we may rotate it by a non-negative integer K so that the array becomes A[K], A[K+1], A{K+2], ... A[A.length - 1], A[0], A[1], ..., A[K-1].  Afterward, any entries that are less than or equal to their index are worth 1 point. 

For example, if we have [2, 4, 1, 3, 0], and we rotate by K = 2, it becomes [1, 3, 0, 2, 4].  This is worth 3 points because 1 > 0 [no points], 3 > 1 [no points], 0 <= 2 [one point], 2 <= 3 [one point], 4 <= 4 [one point].

Over all possible rotations, return the rotation index K that corresponds to the highest score we could receive.  If there are multiple answers, return the smallest such index K.

Example 1:
Input: [2, 3, 1, 4, 0]
Output: 3
Explanation:  
Scores for each K are listed below: 
K = 0,  A = [2,3,1,4,0],    score 2
K = 1,  A = [3,1,4,0,2],    score 3
K = 2,  A = [1,4,0,2,3],    score 3
K = 3,  A = [4,0,2,3,1],    score 4
K = 4,  A = [0,2,3,1,4],    score 3

So we should choose K = 3, which has the highest score.

 

Example 2:
Input: [1, 3, 0, 2, 4]
Output: 0
Explanation:  A will always have 3 points no matter how it shifts.
So we will choose the smallest K, which is 0.

Note:

       A will have length at most 20000.
       A[i] will be in the range [0, A.length].
assistant
#Here is the Python code you requested.

 class Solution:
     def movesToChessboard(self, board):
         N = len(board)
         R = [sum([board[i][j]<<j for j in range(N)]) for i in range(N) ]
         C = [sum([board[i][j]<<i for i in range(N)]) for j in range(N) ]
         def calcMove(R, N):
             def sumOne(n) :
                 k = 0
                 while n:
                     n = n&(n-1)
                     k+=1
                 return k
             d = {}
             for i in range(N):
                 if not R[i] in d:
                     d[R[i]] = []
                 d[R[i]].append(i)
             akey =[]
             for k in d:
                 akey.append(k)
             
             if len(akey)==2 and akey[0]^akey[1] == (1<<N) -1 and abs(sumOne(akey[0])-sumOne(akey[1]))<=1:
                 a0, a1 = d[akey[0]], d[akey[1]]
                 if abs(len(a0)-len(a1))>1:
                     return -1
                 if len(a0) >= len(a1):
                     res = sum( [i%2  for i in a0])
                 else:
                     res = sum( [i%2  for i in a1])
                 return res if N%2 else min(res, max(N//2-res, 0))
             return -1
         r, c = calcMove(R, N), calcMove(C, N)
         return -1 if r == -1 or c == -1 else r + c