class Solution:
# 官方题解，贪心
def minDominoRotations1(self, A: List[int], B: List[int]) -> int:
def check(x):
"""
Return min number of swaps
if one could make all elements in A or B equal to x.
Else return -1.
"""
# how many rotations should be done
# to have all elements in A equal to x
# and to have all elements in B equal to x
rotations_a = rotations_b = 0
for i in range(n):
# rotations coudn't be done
if A[i] != x and B[i] != x:
return -1
# A[i] != x and B[i] == x
elif A[i] != x:
rotations_a += 1
# A[i] == x and B[i] != x
elif B[i] != x:
rotations_b += 1
# min number of rotations to have all
# elements equal to x in A or B
return min(rotations_a, rotations_b)
n = len(A)
rotations = check(A[0])
# If one could make all elements in A or B equal to A[0]
if rotations != -1 or A[0] == B[0]:
return rotations
# If one could make all elements in A or B equal to B[0]
else:
return check(B[0])
def minDominoRotations(self, A: List[int], B: List[int]) -> int:
def get_min_cnt(key: int) -> int:
up = 0 #都放在上面行
down = 0 #都放在下面行
for x,y in zip(A, B):
if x != key and y != key:
return float('inf')
elif x == key and y == key:
continue
elif x == key and y != key:
down += 1
else:
up += 1
return min(up, down)
res1 = get_min_cnt(A[0])
res2 = get_min_cnt(B[0])
return -1 if (res1 == res2 == float('inf')) else min(res1, res2)
func minDominoRotations(A []int, B []int) int {
nums := [7]int{}
as := [7]int{}
bs := [7]int{}
for i := 0; i < len(A); i++ {
if A[i] == B[i] {
nums[A[i]]++
} else {
nums[A[i]]++
as[A[i]]++
nums[B[i]]++
bs[B[i]]++
}
}
for k, v := range nums {
if v == len(A) {
return min(as[k], bs[k])
}
}
return -1
}
func min(x, y int) int {
if x < y {
return x
}
return y
}
class Solution {
public:
/*
Return min number of rotations
if one could make all elements in A or B equal to x.
Else return -1.
*/
int check(int x, vector<int>& A, vector<int>& B, int n) {
// how many rotations should be done
// to have all elements in A equal to x
// and to have all elements in B equal to x
int rotations_a = 0, rotations_b = 0;
for (int i = 0; i < n; i++) {
// rotations coudn't be done
if (A[i] != x && B[i] != x) return -1;
// A[i] != x and B[i] == x
else if (A[i] != x) rotations_a++;
// A[i] == x and B[i] != x
else if (B[i] != x) rotations_b++;
}
// min number of rotations to have all
// elements equal to x in A or B
return min(rotations_a, rotations_b);
}
int minDominoRotations(vector<int>& A, vector<int>& B) {
int n = A.size();
int rotations = check(A[0], B, A, n);
// If one could make all elements in A or B equal to A[0]
if (rotations != -1 || A[0] == B[0]) return rotations;
// If one could make all elements in A or B equal to B[0]
else return check(B[0], B, A, n);
}
};
class Solution {
/*
Return min number of rotations
if one could make all elements in A or B equal to x.
Else return -1.
*/
public int check(int x, int[] A, int[] B, int n) {
// how many rotations should be done
// to have all elements in A equal to x
// and to have all elements in B equal to x
int rotations_a = 0, rotations_b = 0;
for (int i = 0; i < n; i++) {
// rotations coudn't be done
if (A[i] != x && B[i] != x) return -1;
// A[i] != x and B[i] == x
else if (A[i] != x) rotations_a++;
// A[i] == x and B[i] != x
else if (B[i] != x) rotations_b++;
}
// min number of rotations to have all
// elements equal to x in A or B
return Math.min(rotations_a, rotations_b);
}
public int minDominoRotations(int[] A, int[] B) {
int n = A.length;
int rotations = check(A[0], B, A, n);
// If one could make all elements in A or B equal to A[0]
if (rotations != -1 || A[0] == B[0]) return rotations;
// If one could make all elements in A or B equal to B[0]
else return check(B[0], B, A, n);
}
}
