Find closest integer with the same weight

Given a positive integer X the task is to find an integer Y such that:

The count of set bits is Y is equal to the count of set bits in X.
X != Y.
|X – Y| is minimum.

Examples:

Input: X = 92
Output: 90
90 is the closest number to 92 having
equal number of set bits.

Input: X = 17
Output: 18

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: A little math can lead us to the solution approach. Since the number of bits in both the numbers has to be the same, if a set bit is flipped then an unset bit will also have to be flipped.
Now the problem reduces to choosing two bits for the flipping. Suppose one bit at index i is flipped and another bit at index j (j < i) from the LSB (least significant bit). Then the absolute value of the difference between the original integer and the new one is 2ⁱ – 2^j. To minimize this, i has to be as small as possible and j has to be as close to i as possible.
Since the number of set bits have to be equal, so the bit at index i must be different from the bit at index j. This means that the smallest can be the rightmost bit that's different from the LSB, and j must be the very next bit. In summary, the correct approach is to swap the two rightmost consecutive bits that are different.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
const int NumUnsignBits = 64; 
  
// Function to return the number 
// closest to x which has equal 
// number of set bits as x 
unsigned long findNum(unsigned long x) 
{ 
    // Loop for each bit in x and 
    // compare with the next bit 
    for (int i = 0; i < NumUnsignBits - 1; i++) { 
        if (((x >> i) & 1) != ((x >> (i + 1)) & 1)) { 
            x ^= (1 << i) | (1 << (i + 1)); 
            return x; 
        } 
    } 
} 
  
// Driver code 
int main() 
{ 
    int n = 92; 
  
    cout << findNum(n); 
  
    return 0; 
} 

Java

// Java implementation of the approach 
class GFG  
{ 
static int NumUnsignBits = 64; 
  
// Function to return the number 
// closest to x which has equal 
// number of set bits as x 
static long findNum(long x) 
{ 
    // Loop for each bit in x and 
    // compare with the next bit 
    for (int i = 0; i < NumUnsignBits - 1; i++) 
    { 
        if (((x >> i) & 1) != ((x >> (i + 1)) & 1)) 
        { 
            x ^= (1 << i) | (1 << (i + 1)); 
            return x; 
        } 
    } 
    return Long.MIN_VALUE; 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    int n = 92; 
  
    System.out.println(findNum(n)); 
} 
} 
  
// This code is contributed by PrinciRaj1992 

Python3

# Python3 implementation of the approach  
NumUnsignBits = 64;  
  
# Function to return the number  
# closest to x which has equal  
# number of set bits as x  
def findNum(x) :  
  
    # Loop for each bit in x and  
    # compare with the next bit  
    for i in range(NumUnsignBits - 1) : 
        if (((x >> i) & 1) != ((x >> (i + 1)) & 1)) : 
            x ^= (1 << i) | (1 << (i + 1));  
            return x; 
  
# Driver code  
if __name__ == "__main__" :  
    n = 92;  
    print(findNum(n));  
  
# This code is contributed by AnkitRai01 

C#

// C# implementation of the approach 
using System; 
      
class GFG  
{ 
static int NumUnsignBits = 64; 
  
// Function to return the number 
// closest to x which has equal 
// number of set bits as x 
static long findNum(long x) 
{ 
    // Loop for each bit in x and 
    // compare with the next bit 
    for (int i = 0; i < NumUnsignBits - 1; i++) 
    { 
        if (((x >> i) & 1) != ((x >> (i + 1)) & 1)) 
        { 
            x ^= (1 << i) | (1 << (i + 1)); 
            return x; 
        } 
    } 
    return long.MinValue; 
} 
  
// Driver code 
public static void Main(String[] args) 
{ 
    int n = 92; 
  
    Console.WriteLine(findNum(n)); 
} 
} 
  
// This code is contributed by Rajput-Ji 

Output:

Time Complexity: O(logn)

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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