给定范围内的最大按位与对

给定范围[L，R] ，任务是找到一个对(X，Y) ，以使L≤X 并且X＆Y在所有可能的对中最大，然后打印找到的对的按位与。

例子：

Input: L = 1, R = 9
Output: 8
In all the possible pairs, pair (8, 9) gives the maximum value for bitwise AND.

Input: L = 523641, R = 985624
Output: 985622

天真的方法：从L到R进行迭代，并检查每个可能的对的按位与，并在最后打印最大值。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum bitwise AND
// possible among all the possible pairs
int maxAND(int L, int R)
{
    int maximum = L & R;
  
    for (int i = L; i < R; i++)
        for (int j = i + 1; j <= R; j++)
  
            // Maximum among all (i, j) pairs
            maximum = max(maximum, (i & j));
  
    return maximum;
}
  
// Driver code
int main()
{
    int L = 1, R = 632;
    cout << maxAND(L, R);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
  
// Function to return the maximum bitwise AND
// possible among all the possible pairs
static int maxAND(int L, int R)
{
    int maximum = L & R;
  
    for (int i = L; i < R; i++)
        for (int j = i + 1; j <= R; j++)
  
            // Maximum among all (i, j) pairs
            maximum = Math.max(maximum, (i & j));
  
    return maximum;
}
  
// Driver code
public static void main(String[] args) 
{
    int L = 1, R = 632;
    System.out.println(maxAND(L, R));
}
}
  
// This code contributed by Rajput-Ji

Python3

# Python 3 implementation of the approach
  
# Function to return the maximum bitwise AND
# possible among all the possible pairs
def maxAND(L, R):
    maximum = L & R
  
    for i in range(L, R, 1):
        for j in range(i + 1, R + 1, 1):
              
            # Maximum among all (i, j) pairs
            maximum = max(maximum, (i & j))
  
    return maximum
  
# Driver code
if __name__ == '__main__':
    L = 1
    R = 632
    print(maxAND(L, R))
  
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach 
using System;
  
class GFG 
{ 
  
// Function to return the maximum bitwise AND 
// possible among all the possible pairs 
static int maxAND(int L, int R) 
{ 
    int maximum = L & R; 
  
    for (int i = L; i < R; i++) 
        for (int j = i + 1; j <= R; j++) 
  
            // Maximum among all (i, j) pairs 
            maximum = Math.Max(maximum, (i & j)); 
  
    return maximum; 
} 
  
// Driver code 
public static void Main(String[] args) 
{ 
    int L = 1, R = 632; 
    Console.WriteLine(maxAND(L, R)); 
} 
} 
  
// This code has been contributed by 29AjayKumar

输出：

630

时间复杂度： O(n ² )

高效方法：如果我们仔细观察[L，R]范围，则(R)和(R – 1)的AND或(R – 1)和(R – 2)的AND之间可能存在最大AND。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum bitwise AND
// possible among all the possible pairs
int maxAND(int L, int R)
{
  
    // If there is only a single value
    // in the range [L, R]
    if (L == R)
        return L;
  
    // If there are only two values
    // in the range [L, R]
    else if ((R - L) == 1)
        return (R & L);
    else {
        if (((R - 1) & R) > ((R - 2) & (R - 1)))
            return ((R - 1) & R);
        else
            return ((R - 2) & (R - 1));
    }
}
  
// Driver code
int main()
{
    int L = 1, R = 632;
    cout << maxAND(L, R);
  
    return 0;
}

Java

// Java implementation of the approach 
class GfG 
{ 
  
// Function to return the maximum bitwise AND 
// possible among all the possible pairs 
static int maxAND(int L, int R) 
{ 
  
    // If there is only a single value 
    // in the range [L, R] 
    if (L == R) 
        return L; 
  
    // If there are only two values 
    // in the range [L, R] 
    else if ((R - L) == 1) 
        return (R & L); 
    else { 
        if (((R - 1) & R) > ((R - 2) & (R - 1))) 
            return ((R - 1) & R); 
        else
            return ((R - 2) & (R - 1)); 
    } 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    int L = 1, R = 632; 
    System.out.println(maxAND(L, R)); 
}
} 
  
// This code is contributed by
// Prerna Saini

Python3

# Python3 implementation of the approach
  
# Function to return the maximum bitwise AND
# possible among all the possible pairs
def maxAND(L, R):
  
    # If there is only a single value
    # in the range [L, R]
    if (L == R):
        return L;
  
    # If there are only two values
    # in the range [L, R]
    elif ((R - L) == 1):
        return (R & L);
    else:
        if (((R - 1) & R) >
            ((R - 2) & (R - 1))):
            return ((R - 1) & R);
        else:
            return ((R - 2) & (R - 1));
  
# Driver code
L = 1;
R = 632;
print(maxAND(L, R));
  
# This code contributed by PrinciRaj1992 

C＃

// C# implementation of the approach 
using System;
  
class GfG 
{ 
  
    // Function to return the maximum bitwise AND 
    // possible among all the possible pairs 
    static int maxAND(int L, int R) 
    { 
      
        // If there is only a single value 
        // in the range [L, R] 
        if (L == R) 
            return L; 
      
        // If there are only two values 
        // in the range [L, R] 
        else if ((R - L) == 1) 
            return (R & L); 
        else 
        { 
            if (((R - 1) & R) > ((R - 2) & (R - 1))) 
                return ((R - 1) & R); 
            else
                return ((R - 2) & (R - 1)); 
        } 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int L = 1, R = 632; 
        Console.WriteLine(maxAND(L, R)); 
    } 
} 
  
// This code is contributed by Ryuga

的PHP

 (($R - 2) & ($R - 1)))
            return (($R - 1) & $R);
        else
            return (($R - 2) & ($R - 1));
    }
}
  
// Driver code
$L = 1;
$R = 632;
echo maxAND($L, $R);
  
// This code is contributed by mits
?>

输出：

630

时间复杂度： O(1)

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