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 i^{th} (0 to N1) 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 N1) 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<bits/stdc++.h> 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
<?php // PHP code to count the // Minimum bits in A and B function totalFlips( $A , $B , $C , $N ) { $count = 0; for ( $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 // N represent total count of Bits $N = 5; $a = "10100" ; $b = "00010" ; $c = "10011" ; echo totalFlips( $a , $b , $c , $N ); // This code is contributed by nitin mittal. ?> 
Output:
2
Time Complexity: O(N)
Auxiliary space: O(1)
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Improved By : nitin mittal