Open In App

Bitwise AND of all the odd numbers from 1 to N

Improve
Improve
Like Article
Like
Save
Share
Report

Given an integer N, the task is to find the bitwise AND (&) of all the odd integers from the range [1, N].

Examples: 

Input: N = 7 
Output:
(1 & 3 & 5 & 7) = 1

Input: N = 1 
Output:

Naive approach: Starting from 1, bitwise AND all the odd numbers ? N.

Below is the implementation of the above approach: 

C++




// 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




// 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




# 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#




// 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


Javascript




<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>


Output: 

1

 

Time Complexity: O(n)

Auxiliary Space: O(1)

Efficient approach: Bitwise AND with 1 will always give 1 as the result if the integer has a one at the least significant bit (all the odd integers in this case). So, the result will be 1 in all the cases.

Below is the implementation of the above approach: 

C++




// 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




// 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




# 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#




// 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


Javascript




<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>


Output: 

1

 

Time Complexity: O(1)

Auxiliary Space: O(1)



Last Updated : 05 May, 2021
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads