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

<p>We may rotate the <code>i<sup>th</sup></code> domino, so that <code>tops[i]</code> and <code>bottoms[i]</code> swap values.</p>

<p>Return the minimum number of rotations so that all the values in <code>tops</code> are the same, or all the values in <code>bottoms</code> are the same.</p>

<p>If it cannot be done, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/14/domino.png" style="height: 300px; width: 421px;" />
<pre>
<strong>Input:</strong> tops = [2,1,2,4,2,2], bottoms = [5,2,6,2,3,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> tops = [3,5,1,2,3], bottoms = [3,6,3,3,4]
<strong>Output:</strong> -1
<strong>Explanation:</strong> 
In this case, it is not possible to rotate the dominoes to make one row of values equal.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= tops.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>bottoms.length == tops.length</code></li>
	<li><code>1 &lt;= tops[i], bottoms[i] &lt;= 6</code></li>
</ul>

class Solution:
    def minDominoRotations(self, tops: List[int], bottoms: List[int]) -> int:










        moves = [] # holds TOPS min moves and BOTTOMS min moves

        # loop for TOPS
        for num in range(1, 7, 1):
            movesBCount = 0      # Reset count for B moves
            movesTCount = 0      # Reset count for T moves
            count = 0.           # Holds count where neither need to flip
            for i in range(0, len(tops), 1):
                # if num in neithter go to next num
                if tops[i] != num and bottoms[i] != num:
                    break
                # if swap Tops will give num
                elif tops[i] != num and bottoms[i] == num:
                    movesTCount += 1
                # if swap Bottoms will give num
                elif tops[i] == num and bottoms[i] != num:
                    movesBCount += 1
                elif tops[i] == num and bottoms[i] == num:
                    count += 1
            # if count of already num + mvoesCont = len(tops) then possible
            if count + movesTCount + movesBCount == len(tops):
                moves.append(movesTCount)
                moves.append(movesBCount)
            
        if moves:
            return min(moves)
        else:
            return -1 # no solution found
16700