1007. Minimum Domino Rotations For Equal Row

In a row of dominoes, A[i] and B[i] represent the top and bottom halves of the i-th domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)

We may rotate the i-th domino, so that A[i] and B[i] swap values.

Return the minimum number of rotations so that all the values in A are the same, or all the values in B are the same.

If it cannot be done, return -1.

Example 1:

Input: A = [2,1,2,4,2,2], B = [5,2,6,2,3,2]
Output: 2
Explanation: 
The first figure represents the dominoes as given by A and B: before we do any rotations.
If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.

Example 2:

Input: A = [3,5,1,2,3], B = [3,6,3,3,4]
Output: -1
Explanation: 
In this case, it is not possible to rotate the dominoes to make one row of values equal.

Note:

1. 1 <= A[i], B[i] <= 6

2. 2 <= A.length == B.length <= 20000

解题要点：

随机取一个index从A，B里，然后算成使A全变成一样数，和使B全变成一样数哪个的交换次数少。如果不能变成一样，则返回-1。

class Solution(object):
    def minDominoRotations(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: int
        """
        def check(A, B, val):
            outA = outB = 0
            for i in range(len(A)):
                if A[i] != val and B[i] != val:
                    return -1
                elif A[i] != val and B[i] == val:
                    outA += 1
                elif A[i] == val and B[i] != val:
                    outB += 1
            return min(outA, outB)
        
        if check(A, B, A[0]) != -1:
            return check(A, B, A[0])
        else:
            return check(A, B, B[0])