AND bit a bit de todos los números impares del 1 al N

Dado un entero N , la tarea es encontrar el AND bit a bit (&) de todos los enteros impares del rango [1, N] .

Ejemplos: 

Entrada: N = 7 
Salida: 1 
(1 y 3 y 5 y 7) = 1

Entrada: N = 1 
Salida: 1 

Enfoque ingenuo: a partir de 1 , bit a bit Y todos los números impares ≤ N .

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
int bitwiseAndOdd(int n)
{
    // Initialize result to 1
    int result = 1;
 
    // Starting from 3, bitwise AND
    // all the odd integers less
    // than or equal to n
    for (int i = 3; i <= n; i = i + 2) {
        result = (result & i);
    }
    return result;
}
 
// Driver code
int main()
{
    int n = 10;
 
    cout << bitwiseAndOdd(n);
 
    return 0;
}

// Java implementation of the approach
class GFG
{
     
    // Function to return the bitwise AND
    // of all the odd integers from
    // the range [1, n]
    static int bitwiseAndOdd(int n)
    {
        // Initialize result to 1
        int result = 1;
     
        // Starting from 3, bitwise AND
        // all the odd integers less
        // than or equal to n
        for (int i = 3; i <= n; i = i + 2)
        {
            result = (result & i);
        }
        return result;
    }
     
    // Driver code
    public static void main (String[] args)
    {
         
        int n = 10;
         
        System.out.println(bitwiseAndOdd(n));
    }
}
 
// This code is contributed by AnkitRai01

# Python3 implementation of the approach
 
# Function to return the bitwise AND
# of all the odd integers from
# the range [1, n]
def bitwiseAndOdd(n) :
 
    # Initialize result to 1
    result = 1;
 
    # Starting from 3, bitwise AND
    # all the odd integers less
    # than or equal to n
    for i in range(3, n + 1, 2) :
        result = (result & i);
 
    return result;
 
# Driver code
if __name__ == "__main__" :
 
    n = 10;
 
    print(bitwiseAndOdd(n));
 
# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to return the bitwise AND
    // of all the odd integers from
    // the range [1, n]
    static int bitwiseAndOdd(int n)
    {
        // Initialize result to 1
        int result = 1;
     
        // Starting from 3, bitwise AND
        // all the odd integers less
        // than or equal to n
        for (int i = 3; i <= n; i = i + 2)
        {
            result = (result & i);
        }
        return result;
    }
     
    // Driver code
    public static void Main()
    {
         
        int n = 10;
        Console.WriteLine(bitwiseAndOdd(n));
    }
}
 
// This code is contributed by AnkitRai01

<script>
 
// Javascript implementation of the approach
 
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
function bitwiseAndOdd(n)
{
     
    // Initialize result to 1
    var result = 1;
 
    // Starting from 3, bitwise AND
    // all the odd integers less
    // than or equal to n
    for(var i = 3; i <= n; i = i + 2)
    {
        result = (result & i);
    }
    return result;
}
 
// Driver code
var n = 10;
 
document.write(bitwiseAndOdd(n));
 
// This code is contributed by noob2000
 
</script>

1

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1)

Enfoque eficiente: AND bit a bit con 1 siempre dará 1 como resultado si el entero tiene un uno en el bit menos significativo (todos los enteros impares en este caso). Entonces, el resultado será 1 en todos los casos.

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
int bitwiseAndOdd(int n)
{
    return 1;
}
 
// Driver code
int main()
{
    int n = 10;
 
    cout << bitwiseAndOdd(n);
 
    return 0;
}

// Java implementation of the approach
class GFG
{
 
    // Function to return the bitwise AND
    // of all the odd integers from
    // the range [1, n]
    static int bitwiseAndOdd(int n)
    {
        return 1;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 10;
     
        System.out.println(bitwiseAndOdd(n));
    }
}
 
// This code is contributed by AnkitRai01

# Python 3 implementation of the approach
 
# Function to return the bitwise AND
# of all the odd integers from
# the range [1, n]
def bitwiseAndOdd(n):
    return 1
 
# Driver code
n = 10
print(bitwiseAndOdd(n))
 
# This code is contributed by ApurvaRaj

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the bitwise AND
    // of all the odd integers from
    // the range [1, n]
    static int bitwiseAndOdd(int n)
    {
        return 1;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 10;
     
        Console.WriteLine(bitwiseAndOdd(n));
    }
}
 
// This code is contributed by AnkitRai01

<script>
// Javascript implementation of the approach
 
// Function to return the bitwise AND
// of all the odd integers from
// the range [1, n]
function bitwiseAndOdd(n)
{
    return 1;
}
 
// Driver code
var n = 10;
document.write( bitwiseAndOdd(n));
 
//This code is contributed by SoumikMondal
</script>

1

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)