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 <= A[i], B[i] <= 6
2 <= A.length == B.length <= 20000

解题思路

要实现某一面都是相同的点数，只需要判断A[0]和B[0]即可。假设A[0]和B[0]都不能作为这个相同的点数，那么说明通过翻转不能实现某一面都是相同的点数。

实现代码

//Runtime: 5 ms, faster than 60.20% of Java online submissions for Minimum Domino Rotations For Equal Row.
//Memory Usage: 45.5 MB, less than 100.00% of Java online submissions for Minimum Domino Rotations For Equal Row.
class Solution {
    public int minDominoRotations(int[] A, int[] B) {
        for (int i = 0, ra = 0, rb = 0; i < A.length && (A[i] == A[0] || B[i] == A[0]); i++) {
            if (A[i] != A[0]) {
                ++ra;
            }

            if (B[i] != A[0]) {
                ++rb;
            }

            if (i == A.length - 1) {
                return Math.min(ra, rb);
            }
        }

        for (int i = 0, ra = 0, rb = 0; i < A.length && (A[i] == B[0] || B[i] == B[0]); i++) {
            if (A[i] != B[0]) {
                ++ra;
            }

            if (B[i] != B[0]) {
                ++rb;
            }

            if (i == A.length - 1) {
                return Math.min(ra, rb);
            }
        }

        return -1;
    }
}