Largest number less than X having at most K set bits

Given an integer X > 1 and an integer K > 0, the task is to find the greatest odd number < X such that the number of 1’s in its binary representation is at most K.

Examples:

Input: X = 10, K = 2 
Output: 10

Input: X = 29, K = 2 
Output: 24 

Naive Approach: Starting from X – 1 check all the numbers below X which have at most K set bits, the first number satisfying the condition is the required answer.

Efficient Approach: is to count the set bits. If the count is less than or equal to K, return X. Otherwise, keep removing rightmost set bit while count – k does not become 0.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the greatest number <= X
// having at most K set bits.
int greatestKBits(int X, int K)
{
    int set_bit_count = __builtin_popcount(X);
    if (set_bit_count <= K)
        return X;

    // Remove rightmost set bits one
    // by one until we count becomes k
    int diff = set_bit_count - K;
    for (int i = 0; i < diff; i++)
        X &= (X - 1);

    // Return the required number
    return X;
}

// Driver code
int main()
{
    int X = 21, K = 2;
    cout << greatestKBits(X, K);
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

class GFG {

    // Function to return the greatest number <= X
    // having at most K set bits.
     int greatestKBits(int X, int K)
     {
       int set_bit_count = Integer.bitCount(X);
       if (set_bit_count <= K)
            return X;

        // Remove rightmost set bits one
        // by one until we count becomes k
        int diff = set_bit_count - K;
        for (int i = 0; i < diff; i++)
            X &= (X - 1);

        // Return the required number
        return X;
    }

// Driver code
public static void main (String[] args)
{
    int X = 21, K = 2;
    GFG g=new GFG();
      System.out.print(g.greatestKBits(X, K));
}

//This code is contributed by Shivi_Aggarwal
}

Python3

# Python 3 implementation of the approach

# Function to return the greatest
# number <= X having at most K set bits.
def greatestKBits(X, K):
    set_bit_count = bin(X).count('1')
    if (set_bit_count <= K):
        return X

    # Remove rightmost set bits one
    # by one until we count becomes k
    diff = set_bit_count - K
    for i in range(0, diff, 1):
        X &= (X - 1)

    # Return the required number
    return X

# Driver code
if __name__ == '__main__':
    X = 21
    K = 2
    print(greatestKBits(X, K))

# This code is contributed by
# Shashank_Sharma

C#

// C# implementation of the above 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 greatest number <= X
    // having at most K set bits.
    static int greatestKBits(int X, int K)
    {
        int set_bit_count = countSetBits(X);
        if (set_bit_count <= K)
        return X;

        // Remove rightmost set bits one
        // by one until we count becomes k
        int diff = set_bit_count - K;
        for (int i = 0; i < diff; i++)
            X &= (X - 1);

        // Return the required number
        return X;
    }

    // Driver code
    public static void Main()
    {
        int X = 21, K = 2;
        Console.WriteLine(greatestKBits(X, K));

    }
}

// This code is contributed by Ryuga

Javascript

<script>

// Javascript implementation of the above 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 greatest number <= X
// having at most K set bits.
function greatestKBits( X, K)
{
    let set_bit_count = countSetBits(X);
    if (set_bit_count <= K)
    return X;

    // Remove rightmost set bits one
    // by one until we count becomes k
    let diff = set_bit_count - K;
    for (let i = 0; i < diff; i++)
        X &= (X - 1);

    // Return the required number
    return X;
}


// Driver Code

let X = 21, K = 2;
document.write(greatestKBits(X, K));

</script>

Output

20

Time Complexity: O(log₂X), as the time complexity of __builtin_popcount(X) is O(log₂X)
Auxiliary Space: O(1)