给定一个N位的三个二进制序列A，B和C的序列。计算翻转A和B所需的最小位，以使A和B的XOR等于C。例如：
Input: N = 3
A = 110
B = 101
C = 001
Output: 1
We only need to flip the bit of 2nd position
of either A or B, such that A ^ B = C i.e.,
100 ^ 101 = 001天真的方法是生成A和B中所有可能的比特组合，然后对其进行XOR运算以检查其是否等于C。这种方法的时间复杂度呈指数增长，因此对于较大的N值来说，这不会更好。
一种有效的方法是使用XOR的概念。
XOR Truth Table
Input Output
X Y Z
0 0 - 0
0 1 - 1
1 0 - 1
1 1 - 0如果概括一下，我们将发现在A和B的任何位置上，我们只需要翻转A或B的第i个位置(从0到N-1)，否则我们将无法获得最少的位数。
因此，在i的任何位置(0到N-1)，您都会遇到两种情况，即A [i] == B [i]或A [i]！= B [i]。让我们一一讨论。
- 如果A [i] == B [i]，那么这些位的XOR将为0，则在C []中会出现两种情况：C [i] == 0或C [i] == 1。
如果C [i] == 0，则无需翻转位，但如果C [i] == 1，则我们必须翻转A [i]或B [i]中的位，以便1 ^ 0 == 1或0 ^ 1 == 1。 - 如果A [i]！= B [i]，则这些位的XOR值等于1，在C中再次出现两种情况，即C [i] == 0或C [i] == 1。
因此，如果C [i] == 1，则无需翻转位，但如果C [i] == 0，则需要翻转A [i]或B [i]中的位，以便0 ^ 0 == 0或1 ^ 1 == 0
C++
// C++ code to count the Minimum bits in A and B
#include
using namespace std;
int totalFlips(char *A, char *B, char *C, int N)
{
int count = 0;
for (int i=0; i < N; ++i)
{
// If both A[i] and B[i] are equal
if (A[i] == B[i] && C[i] == '1')
++count;
// If Both A and B are unequal
else if (A[i] != B[i] && C[i] == '0')
++count;
}
return count;
}
//Driver Code
int main()
{
//N represent total count of Bits
int N = 5;
char a[] = "10100";
char b[] = "00010";
char c[] = "10011";
cout << totalFlips(a, b, c, N);
return 0;
} Java
// Java code to count the Minimum bits in A and B
class GFG {
static int totalFlips(String A, String B,
String C, int N)
{
int count = 0;
for (int i = 0; i < N; ++i)
{
// If both A[i] and B[i] are equal
if (A.charAt(i) == B.charAt(i) &&
C.charAt(i) == '1')
++count;
// If Both A and B are unequal
else if (A.charAt(i) != B.charAt(i)
&& C.charAt(i) == '0')
++count;
}
return count;
}
//driver code
public static void main (String[] args)
{
//N represent total count of Bits
int N = 5;
String a = "10100";
String b = "00010";
String c = "10011";
System.out.print(totalFlips(a, b, c, N));
}
}
// This code is contributed by Anant Agarwal.Python
# Python code to find minimum bits to be flip
def totalFlips(A, B, C, N):
count = 0
for i in range(N):
# If both A[i] and B[i] are equal
if A[i] == B[i] and C[i] == '1':
count=count+1
# if A[i] and B[i] are unequal
elif A[i] != B[i] and C[i] == '0':
count=count+1
return count
# Driver Code
# N represent total count of Bits
N = 5
a = "10100"
b = "00010"
c = "10011"
print(totalFlips(a, b, c, N))C#
// C# code to count the Minimum
// bits flip in A and B
using System;
class GFG {
static int totalFlips(string A, string B,
string C, int N)
{
int count = 0;
for (int i = 0; i < N; ++i) {
// If both A[i] and B[i] are equal
if (A[i] == B[i] && C[i] == '1')
++count;
// If Both A and B are unequal
else if (A[i] != B[i] && C[i] == '0')
++count;
}
return count;
}
// Driver code
public static void Main()
{
// N represent total count of Bits
int N = 5;
string a = "10100";
string b = "00010";
string c = "10011";
Console.Write(totalFlips(a, b, c, N));
}
}
// This code is contributed by Anant Agarwal.PHP
Javascript
输出：
2