[LeetCode][C++] #1007. Minimum Domino Rotations For Equal Row

[Medium][Question]:

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

We may rotate the ith domino, so that tops[i] and bottoms[i] swap values.

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

If it cannot be done, return -1.

Example 1:

Input: tops = [2,1,2,4,2,2], bottoms = [5,2,6,2,3,2]
Output: 2
Explanation: 
The first figure represents the dominoes as given by tops and bottoms: 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: tops = [3,5,1,2,3], bottoms = [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.

Constraints:

• 2 <= tops.length <= 2 * 104
• bottoms.length == tops.length
• 1 <= tops[i], bottoms[i] <= 6

My Solution[C++]:

[Ideas]: 這題比較考驗思路，因為骰子只有六種點數，所以
Step1. 找出要當作翻轉基準的點數(x)是哪一個，當上排x數量+下排x數量－重複x數量=排的長度，就代表每一格的上下都有該X點數

Step2. 找出最小值，有兩個動作，第一，要選擇往上翻或是往下翻。第二，因為可翻轉的基準點數不止一個，所以要做比較