Count minimum bits to flip such that XOR of A and B equal to C

• Difficulty Level : Easy
• Last Updated : 25 Jan, 2022

Given a sequence of three binary sequences A, B and C of N bits. Count the minimum bits required to flip in A and B such that XOR of A and B is equal to C. For Example :

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

A Naive approach is to generate all possible combination of bits in A and B and then XORing them to Check whether it is equal to C or not. Time complexity of this approach grows exponentially so it would not be better for large value of N.
An Efficient approach is to use concept of XOR.

XOR Truth Table
Input    Output
X     Y     Z
0     0  -  0
0     1  -  1
1     0  -  1
1     1  -  0

If we generalize, we will find that at any position of A and B, we just only need to flip ith (0 to N-1) position of either A or B otherwise we will not able to achieve minimum no of Bits.
So at any position of i (0 to N-1) you will encounter two type of situation i.e., either A[i] == B[i] or A[i] != B[i]. Let’s discuss it one by one.

• If A[i] == B[i] then XOR of these bits will be 0, two cases arise in C[]: C[i]==0 or C[i]==1.
If C[i] == 0, then no need to flip the bit but if C[i] == 1 then we have to flip the bit either in A[i] or B[i] so that 1^0 == 1 or 0^1 == 1.

• If A[i] != B[i] then XOR of these Bits gives a 1, In C two cases again arise i.e., either C[i] == 0 or C[i] == 1.
Therefore if C[i] == 1, then we need not to flip the bit but if C[i] == 0, then we need to flip the bit either in A[i] or B[i] so that 0^0==0 or 1^1==0

C++

 // C++ code to count the Minimum bits in A and B#includeusing 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 Codeint 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 Bclass 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.

Python3

 # Python code to find minimum bits to be flipdef 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 BitsN = 5a = "10100"b = "00010"c = "10011"print(totalFlips(a, b, c, N))

C#

 // C# code to count the Minimum// bits flip in A and Busing 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.



Javascript



Output:

2

Time Complexity: O(N)
Auxiliary space: O(1)
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