In a row of dominoes, A[i] and B[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 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.
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.
Constraints:
2 <= A.length == B.length <= 2 * 104
1 <= A[i], B[i] <= 6
Solutions
🐍 Python
classSolution:defminDominoRotations(self,A: List[int],B: List[int]) ->int: cA, cB =Counter(A).most_common(),Counter(B).most_common()#find most frequent on top and bottom top_frequent, bottom_frequent = [cA[0][0]], [cB[0][0]]#if there more then one element with top count, add them top_frequent.extend([ch for ch,i in cA[1:] if i == cA[0][0]]) bottom_frequent.extend([ch for ch,i in cB[1:] if i == cB[0][0]])#find min rotation (try top then bottom) min_count =-1for ch in top_frequent: local_counter =0for i, domino_ch inenumerate(A):if domino_ch == ch:continueelif B[i]== ch: local_counter+=1continueelse:breakelse:if min_count ==-1or local_counter < min_count: min_count = local_counterfor ch in bottom_frequent: local_counter =0for i, domino_ch inenumerate(B):if domino_ch == ch:continueelif A[i]== ch: local_counter+=1continueelse:breakelse:if min_count ==-1or local_counter < min_count: min_count = local_counterreturn min_count