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

• Difficulty Level : Easy
• Last Updated : 16 May, 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.
Another 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.`

## Python3

 `# 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``        ``else` `if` `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

 ``
Output
`2`

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

Efficient Approach:

This approach follows O(log N) time complexity.

## C++

 `// C++ code to count the Minimum bits in A and B``#include ``using` `namespace` `std;` `int` `totalFlips(string A, string B, string C, ``int` `N)``{``    ``int` `INTSIZE = 31;``    ``int` `ans = 0;``    ``int` `i = 0;``    ``while` `(N > 0) {``        ``// Considering only 31 bits``        ``int` `a = stoi(A.substr(i * INTSIZE, min(INTSIZE, N)),``                     ``0, 2);``        ``int` `b = stoi(B.substr(i * INTSIZE, min(INTSIZE, N)),``                     ``0, 2);``        ``int` `c = stoi(C.substr(i * INTSIZE, min(INTSIZE, N)),``                     ``0, 2);``        ``int` `Z = a ^ b ^ c;``        ``// builtin function for``        ``// counting the number of set bits.``        ``ans += __builtin_popcount(Z);``        ``i++;``        ``N -= 32;``    ``}` `    ``return` `ans;``}` `// 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;``}` `// This code is contributed by Kasina Dheeraj.`

## 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` `INTSIZE = ``31``;``        ``int` `ans = ``0``;``        ``int` `i = ``0``;``        ``while` `(N > ``0``) {``            ``// Considering only 31 bits``            ``int` `a = Integer.parseInt(``                ``A.substring(i * INTSIZE,``                            ``i * INTSIZE``                                ``+ Math.min(INTSIZE, N)),``                ``2``);``            ``int` `b = Integer.parseInt(``                ``B.substring(i * INTSIZE,``                            ``i * INTSIZE``                                ``+ Math.min(INTSIZE, N)),``                ``2``);``            ``int` `c = Integer.parseInt(``                ``C.substring(i * INTSIZE,``                            ``i * INTSIZE``                                ``+ Math.min(INTSIZE, N)),``                ``2``);``            ``int` `Z = a ^ b ^ c;``            ``// builtin function for``            ``// counting the number of set bits.``            ``ans += Integer.bitCount(Z);``            ``i++;``            ``N -= ``32``;``        ``}` `        ``return` `ans;``    ``}` `    ``// 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 Kasina Dheeraj.`
Output
`2`

Why this code works?

We observe that bit must be flipped if A[i]^B[i] !=C[i]. So, we can get the number of flips by calculating the number of set bits in a^b^c where a,b,c are integer representations of binary string. But string length may be greater than 32, size of a typical int type. So, the plan is to divide the string into substrings of length 31 ,perform operations and count set bits as mentioned for each substring.

Time Complexity: O(log N) as the while loop runs for log31N times and counting set bits account for at most O(32) for 32-bit and O(64) for 64-bit and for each substring operation O(31).

Space Complexity: O(1) , to be noted that substring operation need O(32) space.

This article is contributed by Shubham Bansal and Kasina Dheeraj . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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