小于X的最大数字，最多具有K个设置位

给定整数X> 1和整数K> 0 ，任务是找到最大的奇数 ，以使其二进制表示中的1的个数最多为K。

例子：

Input: X = 10, K = 2
Output: 10

Input: X = 29, K = 2
Output: 24

幼稚的方法：从X – 1开始，检查X以下所有最多具有K个设置位的数字，满足条件的第一个数字是必需的答案。

高效方法：是对设置的位进行计数。如果计数小于或等于K，则返回X。否则，在计数– k不会变为0的同时，继续删除最严格的设置位。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

输出：

20