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

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  • Difficulty Level : Hard
  • Last Updated : 28 Jul, 2022
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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:
 

 

Recommended Practice

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 <bits/stdc++.h>
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<<n;
            n++;
            setBits--;
        }
         
        return res;
    }
 
// Driver code
int main()
{
    int a = 3, b = 5;
 
    cout << minVal(a, b);
 
    return 0;
}

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
        i = 0
    # we need to check for remaining set bits
    # we can assign remaining set bits to least significant bits
    while(setBits > 0):
        mask = (1 << i)
        if (ans & mask) == 0:
            ans |= (1 << i)
            setBits -= 1
        i += 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




<script>
// Javascript implementation of the approach
 
// Function to get no of set
// bits in binary representation
// of positive integer n
function countSetBits(n) {
    let 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
function minVal(a, b)
{
 
    // Count of set-bits in bit
    let setBits = countSetBits(b);
    let ans = 0;
 
    for (let i = 30; i >= 0; i--) {
        let 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
let a = 3, b = 5;
 
document.write(minVal(a, b));
 
// This code is contributed by gfgking.
</script>

Output

3

Time Complexity: O(log(N))

Auxiliary Space: O(log(N))

 2. Space efficient approach

By bit manipulation theory the Xor of two values will be minimised if they have same bits. By using this theory we will be dividing this problem  into 3 parts as follows :

One is if the set bits of A and B are equal. Second is if the the set bits of A are greater than B. Third is if the set bits of B are greater than A. If the set bits are equal then we will return A itself. If it falls under second case then we will remove the difference number of lower bits from A. If it falls under the third case then we will add the lower bits of A if the bit is not set in A.

Complexity
Time Complexity we need to iterate to all the bits present in A and B. The maximum number of bits are 30. so the complexity is O(log(max(A,B))).
Space Complexity We need to  just iterate over the 30 bits and store the answer so the complexity is O(1).

Below is the implementation of the above approach: 

C++




#include <bits/stdc++.h>
using namespace std;
 
int minVal(int a, int b) {
        // code here
        int setb= __builtin_popcount(b);
        int seta= __builtin_popcount(a);
        int ans=0;
        for(int i=0;i<=31;i++){
            int mask=1<<i;
            int set= a & mask;
            if(set==0 && setb>seta){
                ans |=(mask);
                setb--;
            }
            else if(set && seta>setb) seta--;
            else ans|=set;
        }
        return ans;
}
int main()
{
    int a = 7, b = 12;
 
    cout << minVal(a, b);
 
    return 0;
}

Python3




# python3 code for the above approach:
def minVal(a, b):
   
    # code here
    setb = bin(b)[2:].count('1')
    seta = bin(a)[2:].count('1')
    ans = 0
    for i in range(0, 32):
        mask = 1 << i
        set = a & mask
        if(set == 0 and setb > seta):
            ans |= (mask)
            setb -= 1
 
        elif(set and seta > setb):
            seta -= 1
        else:
            ans |= set
 
    return ans
 
 
if __name__ == "__main__":
 
    a = 7
    b = 12
 
    print(minVal(a, b))
 
  # This code is contributed by rakeshsahni

Output

6

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