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 TableInputOutputX 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 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<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.` `?>` |

## Javascript

`<script>` `// Javascript code to count the Minimum bits in A and B` ` ` `function` `totalFlips(A, B, C, N)` ` ` `{` ` ` `let count = 0;` ` ` `for` `(let 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` ` ` `let N = 5;` ` ` `let a = ` `"10100"` `;` ` ` `let b = ` `"00010"` `;` ` ` `let c = ` `"10011"` `;` ` ` ` ` `document.write(totalFlips(a, b, c, N)); ` ` ` `</script>` |

**Output:**

2

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

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