1007. Minimum Domino Rotations For Equal Row Time: Space: C++Python 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22class Solution { public: int minDominoRotations(vector<int>& A, vector<int>& B) { const int n = A.size(); vector<int> countA(7); vector<int> countB(7); vector<int> countBoth(7); for (int i = 0; i < n; ++i) { ++countA[A[i]]; ++countB[B[i]]; if (A[i] == B[i]) ++countBoth[A[i]]; } for (int i = 1; i <= 6; ++i) if (countA[i] + countB[i] - countBoth[i] == n) return n - max(countA[i], countB[i]); return -1; } }; 1 2 3 4 5 6class Solution: def minDominoRotations(self, A: List[int], B: List[int]) -> int: for num in range(1, 7): if all(num in pair for pair in zip(A, B)): return len(A) - max(A.count(num), B.count(num)) return -1