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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `#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

 ` `

Output:

```2
```

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

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Improved By : nitin mittal

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