# Find a number X such that (X XOR A) is minimum and the count of set bits in X and B are equal

• Difficulty Level : Hard
• Last Updated : 01 Mar, 2022

Given two integers A and B, the task is to find an integer X such that (X XOR A) is minimum possible and the count of set bit in X is equal to the count of set bits in B.
Examples:

Input: A = 3, B = 5
Output:
Binary(A) = Binary(3) = 011
Binary(B) = Binary(5) = 101
The XOR will be minimum when M = 3
i.e. (3 XOR 3) = 0 and the number
of set bits in 3 is equal
to the number of set bits in 5.
Input: A = 7, B = 12
Output:

Approach: It is known that the xor of an element with itself is 0. So, try to generate M’s binary representation as close to A as possible. Traverse from the most significant bit in A to the least significant bit and if a bit is set at the current position then it also needs to be set in the required number in order to minimize the XOR but the number of bits set has to be equal to the number of set bits in B. So, when the count of set bits in the required number has reached the count of set bits in B then the rest of the bits have to be 0.

It can also be possible that the number of set bits in B is more than the number of set bits in A, In this case, start filling the unset bits to set bits from the least significant bit to the most significant bit.

If the number of set bits is still not equal to B then add the remaining number of set bits to the left of the most significant bit in order to make set bits of M equal to the set bits of B.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the value x` `// such that (x XOR a) is minimum` `// and the number of set bits in x` `// is equal to the number` `// of set bits in b` `int` `minVal(``int` `a, ``int` `b) {` `        `  `        ``int` `setBits=0,res=0;` `        `  `          ``//To count number of set bits in b` `        ``setBits= __builtin_popcount(b);` `        `  `        ``//creating binary representation of a in stack s` `        ``stack<``short` `int``> s;` `        ``while``(a>0)` `        ``{` `            ``s.push(a%2);` `            ``a=a/2;` `        ``}` `        `    `        ``// Decrease the count of setBits` `        ``// as in the required number set bits has to be` `        ``// equal to the count of set bits in b` `        ``//Creating nearest possible number in m in binary form.` `        ``//Using vector as the number in binary for can be large.` `        ``vector<``short` `int``> m;    ` `        `  `        ``while``(!s.empty())` `        ``{` `            ``if``(s.top()==1 && setBits>0)` `            ``{` `                ``m.push_back(1);` `                ``setBits--;` `            ``}` `            ``else` `            ``{` `                ``m.push_back(0);` `            ``}` `            ``s.pop();` `        ``}` `        `  `        ``//Filling the unset bits from the least significant bit to the most significant bit ` `        ``//if the setBits are not equal to zero` `        ``for``(``int` `i=m.size()-1;i>=0 && setBits>0;i--)` `        ``{` `            ``if``(m[i]==0)` `            ``{` `                ``m[i]=1;` `                ``setBits--;` `            ``}` `        ``}` `        `  `        `  `        ``int` `mask;` `        ``for``(``int` `i=m.size()-1;i>=0;i--)` `        ``{` `            ``mask=1<<(m.size()-i-1);` `            `  `            ``res+=m[i]*mask;` `        ``}` `        ``int` `n=m.size();` `        `  `        ``//if the number of setBits is still not equal to zero` `        ``//dd the remaining number of set bits to the left of the most significant bit ` `        ``//in order to make set bits of m equal to the set bits of B.` `        `  `        ``while``(setBits>0)` `        ``{` `            ``res+=1<

## Java

 `// Java implementation of the approach` `class` `GFG` `{ ` `    ``// Function to get no of set ` `    ``// bits in binary representation ` `    ``// of positive integer n ` `    ``static` `int` `countSetBits(``int` `n) ` `    ``{ ` `        ``int` `count = ``0``; ` `        ``while` `(n > ``0``) ` `        ``{ ` `            ``count += n & ``1``; ` `            ``n >>= ``1``; ` `        ``} ` `        ``return` `count; ` `    ``} `   `// Function to return the value x` `// such that (x XOR a) is minimum` `// and the number of set bits in x` `// is equal to the number` `// of set bits in b` `static` `int` `minVal(``int` `a, ``int` `b)` `{` `    ``// Count of set-bits in bit` `    ``int` `setBits = countSetBits(b);` `    ``int` `ans = ``0``;`   `    ``for` `(``int` `i = ``30``; i >= ``0``; i--) ` `    ``{` `        ``int` `mask = ``1` `<< i;` `        `  `        ``// If i'th bit is set also set the` `        ``// same bit in the required number` `        ``if` `((a & mask) > ``0` `&& setBits > ``0``) ` `        ``{` `            ``ans |= (``1` `<< i);` `            `  `            ``// Decrease the count of setbits` `            ``// in b as the count of set bits` `            ``// in the required number has to be` `            ``// equal to the count of set bits in b` `            ``setBits--;` `        ``}` `    ``}` `    ``return` `ans;` `}`   `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `a = ``3``, b = ``5``;`   `    ``System.out.println(minVal(a, b));` `} ` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the approach `   `# Function to return the value x ` `# such that (x XOR a) is minimum ` `# and the number of set bits in x ` `# is equal to the number ` `# of set bits in b ` `def` `minVal(a, b) : `   `    ``# Count of set-bits in bit ` `    ``setBits ``=` `bin``(b).count(``'1'``); ` `    ``ans ``=` `0``; `   `    ``for` `i ``in` `range``(``30``, ``-``1``, ``-``1``) :` `        ``mask ``=` `(``1` `<< i); ` `        ``s ``=` `(a & mask); `   `        ``# If i'th bit is set also set the ` `        ``# same bit in the required number ` `        ``if` `(s ``and` `setBits > ``0``) :` `            ``ans |``=` `(``1` `<< i); `   `            ``# Decrease the count of setbits ` `            ``# in b as the count of set bits ` `            ``# in the required number has to be ` `            ``# equal to the count of set bits in b ` `            ``setBits ``-``=` `1``; `   `    ``return` `ans; `   `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: `   `    ``a ``=` `3``; b ``=` `5``; `   `    ``print``(minVal(a, b)); `   `# This code is contributed by kanugargng`

## C#

 `// C# implementation of the approach` `using` `System; `   `class` `GFG` `{ ` `    ``// Function to get no of set ` `    ``// bits in binary representation ` `    ``// of positive integer n ` `    ``static` `int` `countSetBits(``int` `n) ` `    ``{ ` `        ``int` `count = 0; ` `        ``while` `(n > 0) ` `        ``{ ` `            ``count += n & 1; ` `            ``n >>= 1; ` `        ``} ` `        ``return` `count; ` `    ``} `   `// Function to return the value x` `// such that (x XOR a) is minimum` `// and the number of set bits in x` `// is equal to the number` `// of set bits in b` `static` `int` `minVal(``int` `a, ``int` `b)` `{` `    ``// Count of set-bits in bit` `    ``int` `setBits = countSetBits(b);` `    ``int` `ans = 0;`   `    ``for` `(``int` `i = 30; i >= 0; i--) ` `    ``{` `        ``int` `mask = 1 << i;` `        `  `        ``// If i'th bit is set also set the` `        ``// same bit in the required number` `        ``if` `((a & mask) > 0 && setBits > 0) ` `        ``{` `            `  `            ``ans |= (1 << i);` `            `  `            ``// Decrease the count of setbits` `            ``// in b as the count of set bits` `            ``// in the required number has to be` `            ``// equal to the count of set bits in b` `            ``setBits--;` `        ``}` `    ``}`   `    ``return` `ans;` `}`   `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `a = 3, b = 5;`   `    ``Console.Write(minVal(a, b));` `} ` `} `   `// This code is contributed by Mohit kumar 29`

## Javascript

 ``

Output:

`3`

Time Complexity: O(log(N))

Auxiliary Space: O(log(N))

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