# 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

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 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.` |

## 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

`<?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)

**Efficient Approach:**

This approach follows O(log N) time complexity.

## C++

`// C++ code to count the Minimum bits in A and B` `#include <bits/stdc++.h>` `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 log _{31}N 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.”