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.
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.
Constraints:
2 <= A.length == B.length <= 2 * 104
1 <= A[i], B[i] <= 6
Solutions
🐍 Python
class Solution:
def minDominoRotations(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 = -1
for ch in top_frequent:
local_counter = 0
for i, domino_ch in enumerate(A):
if domino_ch == ch:
continue
elif B[i] == ch:
local_counter+=1
continue
else:
break
else:
if min_count == -1 or local_counter < min_count:
min_count = local_counter
for ch in bottom_frequent:
local_counter = 0
for i, domino_ch in enumerate(B):
if domino_ch == ch:
continue
elif A[i] == ch:
local_counter+=1
continue
else:
break
else:
if min_count == -1 or local_counter < min_count:
min_count = local_counter
return min_count
