# 1007. Minimum Domino Rotations For Equal Row

#### Medium

***

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:**

![](https://assets.leetcode.com/uploads/2021/05/14/domino.png)

```
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`

```python
class Solution:
    def minDominoRotations(self, tops: List[int], bottoms: List[int]) -> int:
        total = len(tops)
        sorted_top_counter, top_counter = self.get_best(tops)
        sorted_bottom_counter, bottom_counter = self.get_best(bottoms)
        
        top_diff = float("inf")
        for key, count in sorted_top_counter:
            if total - count < top_diff and (count + bottom_counter[key] >= total) and self.valid(tops, bottoms, key):
                top_diff = total - count
                
        for key, count in sorted_bottom_counter:
            if total - count < top_diff and (count + top_counter[key] >= total) and self.valid(tops, bottoms, key):
                top_diff = total - count
        minimum = top_diff
        return -1 if minimum == float("inf") else minimum
            
    # Check if there exists position for a key in either top or bottom row
    def valid(self, tops, bottoms, key):
        for index in range(len(tops)):
            if key != tops[index] and key != bottoms[index]:
                return False
        return True
        
    # Return counter of keys
    def get_best(self, tops: List[int]):
        top_counter = Counter()
        for top in tops:
            top_counter[top] += 1
        return top_counter.most_common(), top_counter
```

#### Solution 2 Taken from Leetcode

```
class Solution:
    def minDominoRotations(self, A: List[int], B: List[int]) -> int:
        same, countA, countB = Counter(), Counter(A), Counter(B)
        for a, b in zip(A, B):
            if a == b:
                same[a] += 1
        for i in range(1, 7):
            if countA[i] + countB[i] - same[i] == len(A):
                # countA-same or countB-same since `same` is common here take min of two counts
                return min(countA[i], countB[i]) - same[i]        
        return -1
```