Minimize bits to be flipped in X and Y such that their Bitwise OR is equal to Z

Given three positive integers **X**, **Y** and **Z**, the task is to count the minimum numbers of bits required to be flipped in **X** and **Y** such that **X OR Y (X | Y)** is equal to **Z**.**Examples:**

Input :X = 5 Y = 8 Z = 6Output :3Explanation :BeforeAfterFlippingFlippingX :: 01010100Y ::10000010 ------ ----- Z :: 0110 0110 So we need to flip 3 bits in order to make (X | Y) = Z .Input :X = 1 Y = 2 Z = 3Output :0

**Approach :**

In order to solve this problem, we need to observe that the need to flip a bit X or Y arises only for the following conditions:

- If the current bit in Z is set and the corresponding bit in both X and Y is not set, we need to set a bit in either X or in Y.
- If the current bit in Z is unset, we need to unset the corresponding bit in A and B, whichever is set. Unset both if both are set.

Thus, we need to traverse through the bits of X, Y, and Z and follow the above two steps to get the desired result.

Below is the implementation of the above approach:

## C++

`// C++ Program to minimize` `// number of bits to be fipped ` `// in X or Y such that their ` `// bitwise or is equal to minimize` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// This function returns minimum ` `// number of bits to be flipped in` `// X and Y to make X | Y = Z` `int` `minimumFlips(` `int` `X, ` `int` `Y, ` `int` `Z)` `{` ` ` `int` `res = 0;` ` ` `while` `(X > 0 || Y > 0 || Z > 0) {` ` ` ` ` `// If current bit in Z is set and is ` ` ` `// also set in either of X or Y or both` ` ` `if` `(((X & 1) || (Y & 1)) && (Z & 1)) {` ` ` `X = X >> 1;` ` ` `Y = Y >> 1;` ` ` `Z = Z >> 1;` ` ` `continue` `;` ` ` `}` ` ` `// If current bit in Z is set ` ` ` `// and is unset in both X and Y` ` ` `else` `if` `(!(X & 1) && !(Y & 1) && (Z & 1)) {` ` ` `// Set that bit in either X or Y` ` ` `res++;` ` ` `}` ` ` `else` `if` `((X & 1) || (Y & 1) == 1) ` ` ` `{` ` ` `// If current bit in Z is unset ` ` ` `// and is set in both X and Y` ` ` `if` `((X & 1) && (Y & 1) && !(Z & 1)) {` ` ` `// Unset the bit in both X and Y` ` ` `res += 2;` ` ` `}` ` ` `// If current bit in Z is unset and ` ` ` `// is set in either X or Y ` ` ` `else` `if` `(((X & 1) || (Y & 1)) && !(Z & 1)) ` ` ` `{` ` ` `// Unset that set bit` ` ` `res++;` ` ` `}` ` ` `}` ` ` `X = X >> 1;` ` ` `Y = Y >> 1;` ` ` `Z = Z >> 1;` ` ` `}` ` ` `return` `res;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `X = 5, Y = 8, Z = 6;` ` ` `cout << minimumFlips(X, Y, Z);` ` ` `return` `0;` `}` |

*chevron_right*

*filter_none*

## Java

`// Java program to minimize` `// number of bits to be fipped ` `// in X or Y such that their ` `// bitwise or is equal to minimize` `import` `java.util.*;` `class` `GFG{` `// This function returns minimum ` `// number of bits to be flipped in` `// X and Y to make X | Y = Z` `static` `int` `minimumFlips(` `int` `X, ` `int` `Y, ` `int` `Z)` `{` ` ` `int` `res = ` `0` `;` ` ` `while` `(X > ` `0` `|| Y > ` `0` `|| Z > ` `0` `)` ` ` `{` ` ` ` ` `// If current bit in Z is set and is ` ` ` `// also set in either of X or Y or both` ` ` `if` `(((X % ` `2` `== ` `1` `) || ` ` ` `(Y % ` `2` `== ` `1` `)) && ` ` ` `(Z % ` `2` `== ` `1` `))` ` ` `{` ` ` `X = X >> ` `1` `;` ` ` `Y = Y >> ` `1` `;` ` ` `Z = Z >> ` `1` `;` ` ` `continue` `;` ` ` `}` ` ` ` ` `// If current bit in Z is set ` ` ` `// and is unset in both X and Y` ` ` `else` `if` `(!(X % ` `2` `== ` `1` `) &&` ` ` `!(Y % ` `2` `== ` `1` `) &&` ` ` `(Z % ` `2` `== ` `1` `)) ` ` ` `{` ` ` ` ` `// Set that bit in either X or Y` ` ` `res++;` ` ` `}` ` ` `else` `if` `((X % ` `2` `== ` `1` `) || (Y % ` `2` `== ` `1` `)) ` ` ` `{` ` ` ` ` `// If current bit in Z is unset ` ` ` `// and is set in both X and Y` ` ` `if` `((X % ` `2` `== ` `1` `) && ` ` ` `(Y % ` `2` `== ` `1` `) && ` ` ` `!(Z % ` `2` `== ` `1` `))` ` ` `{` ` ` ` ` `// Unset the bit in both X and Y` ` ` `res += ` `2` `;` ` ` `}` ` ` ` ` `// If current bit in Z is unset and ` ` ` `// is set in either X or Y ` ` ` `else` `if` `(((X % ` `2` `== ` `1` `) || ` ` ` `(Y % ` `2` `== ` `1` `)) && ` ` ` `!(Z % ` `2` `== ` `1` `)) ` ` ` `{` ` ` ` ` `// Unset that set bit` ` ` `res++;` ` ` `}` ` ` `}` ` ` `X = X >> ` `1` `;` ` ` `Y = Y >> ` `1` `;` ` ` `Z = Z >> ` `1` `;` ` ` `}` ` ` `return` `res;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `X = ` `5` `, Y = ` `8` `, Z = ` `6` `;` ` ` ` ` `System.out.print(minimumFlips(X, Y, Z));` `}` `}` `// This code is contributed by amal kumar choubey` |

*chevron_right*

*filter_none*

## Python3

`# Python 3 Program to minimize` `# number of bits to be fipped` `# in X or Y such that their` `# bitwise or is equal to minimize` `# This function returns minimum` `# number of bits to be flipped in` `# X and Y to make X | Y = Z` `def` `minimumFlips(X, Y, Z):` ` ` `res ` `=` `0` ` ` `while` `(X > ` `0` `or` `Y > ` `0` `or` ` ` `Z > ` `0` `):` ` ` `# If current bit in Z is ` ` ` `# set and is also set in ` ` ` `# either of X or Y or both` ` ` `if` `(((X & ` `1` `) ` `or` `(Y & ` `1` `)) ` `and` ` ` `(Z & ` `1` `)):` ` ` `X ` `=` `X >> ` `1` ` ` `Y ` `=` `Y >> ` `1` ` ` `Z ` `=` `Z >> ` `1` ` ` `continue` ` ` `# If current bit in Z is set` ` ` `# and is unset in both X and Y` ` ` `elif` `(` `not` `(X & ` `1` `) ` `and` `not` `(Y & ` `1` `) ` `and` ` ` `(Z & ` `1` `)):` ` ` ` ` `# Set that bit in either ` ` ` `# X or Y` ` ` `res ` `+` `=` `1` ` ` `elif` `((X & ` `1` `) ` `or` `(Y & ` `1` `) ` `=` `=` `1` `):` ` ` `# If current bit in Z is unset` ` ` `# and is set in both X and Y` ` ` `if` `((X & ` `1` `) ` `and` `(Y & ` `1` `) ` `and` ` ` `not` `(Z & ` `1` `)):` ` ` ` ` `# Unset the bit in both ` ` ` `# X and Y` ` ` `res ` `+` `=` `2` ` ` `# If current bit in Z is ` ` ` `# unset and is set in ` ` ` `# either X or Y` ` ` `elif` `(((X & ` `1` `) ` `or` `(Y & ` `1` `)) ` `and` ` ` `not` `(Z & ` `1` `)):` ` ` `# Unset that set bit` ` ` `res ` `+` `=` `1` ` ` `X ` `=` `X >> ` `1` ` ` `Y ` `=` `Y >> ` `1` ` ` `Z ` `=` `Z >> ` `1` ` ` `return` `res` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `X ` `=` `5` ` ` `Y ` `=` `8` ` ` `Z ` `=` `6` ` ` `print` `(minimumFlips(X, Y, Z))` `# This code is contributed by Chitranayal` |

*chevron_right*

*filter_none*

## C#

`// C# program to minimize` `// number of bits to be fipped ` `// in X or Y such that their ` `// bitwise or is equal to minimize` `using` `System;` `class` `GFG{` `// This function returns minimum ` `// number of bits to be flipped in` `// X and Y to make X | Y = Z` `static` `int` `minimumFlips(` `int` `X, ` `int` `Y, ` `int` `Z)` `{` ` ` `int` `res = 0;` ` ` `while` `(X > 0 || Y > 0 || Z > 0)` ` ` `{` ` ` ` ` `// If current bit in Z is set and is ` ` ` `// also set in either of X or Y or both` ` ` `if` `(((X % 2 == 1) || ` ` ` `(Y % 2 == 1)) && ` ` ` `(Z % 2 == 1))` ` ` `{` ` ` `X = X >> 1;` ` ` `Y = Y >> 1;` ` ` `Z = Z >> 1;` ` ` `continue` `;` ` ` `}` ` ` ` ` `// If current bit in Z is set ` ` ` `// and is unset in both X and Y` ` ` `else` `if` `(!(X % 2 == 1) &&` ` ` `!(Y % 2 == 1) &&` ` ` `(Z % 2 == 1)) ` ` ` `{` ` ` ` ` `// Set that bit in either X or Y` ` ` `res++;` ` ` `}` ` ` `else` `if` `((X % 2 == 1) || (Y % 2 == 1)) ` ` ` `{` ` ` ` ` `// If current bit in Z is unset ` ` ` `// and is set in both X and Y` ` ` `if` `((X % 2 == 1) && ` ` ` `(Y % 2 == 1) && ` ` ` `!(Z % 2 == 1))` ` ` `{` ` ` ` ` `// Unset the bit in both X and Y` ` ` `res += 2;` ` ` `}` ` ` ` ` `// If current bit in Z is unset and ` ` ` `// is set in either X or Y ` ` ` `else` `if` `(((X % 2 == 1) || ` ` ` `(Y % 2 == 1)) && ` ` ` `!(Z % 2 == 1)) ` ` ` `{` ` ` ` ` `// Unset that set bit` ` ` `res++;` ` ` `}` ` ` `}` ` ` `X = X >> 1;` ` ` `Y = Y >> 1;` ` ` `Z = Z >> 1;` ` ` `}` ` ` `return` `res;` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `int` `X = 5, Y = 8, Z = 6;` ` ` ` ` `Console.Write(minimumFlips(X, Y, Z));` `}` `}` `// This code is contributed by Code_Mech` |

*chevron_right*

*filter_none*

**Output:**

3