Bitwise AND of all even number up to N
Last Updated :
28 Dec, 2022
Given an integer N, the task is to find bitwise and (&) of all even numbers from 1 to N.
Examples:
Input: 2
Output: 2
Input :10
Output : 0
Explanation: Bitwise and of 2, 4, 6, 8 and 10 are 0.
Naive approach: Initialize result as 2. Iterate the loop from 4 to n (for all even numbers) and update the result by finding bitwise and (&).
Below is the implementation of the approach:
C++
// C++ implementation of the above approach
#include <iostream>
using namespace std;
// Function to return the bitwise &
// of all the even numbers upto N
int bitwiseAndTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2) {
result = result & i;
}
return result;
}
// Driver code
int main()
{
int n = 2;
cout << bitwiseAndTillN(n);
return 0;
}
// Java implementation of the above approach
class GFG
{
// Function to return the bitwise &
// of all the even numbers upto N
static int bitwiseAndTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2)
{
result = result & i;
}
return result;
}
// Driver code
public static void main (String[] args)
{
int n = 2;
System.out.println(bitwiseAndTillN(n));
}
}
// This code is contributed by AnkitRai01
# Python3 implementation of the above approach
# Function to return the bitwise &
# of all the even numbers upto N
def bitwiseAndTillN(n) :
# Initialize result as 2
result = 2;
for i in range(4, n + 1, 2) :
result = result & i;
return result;
# Driver code
if __name__ == "__main__" :
n = 2;
print(bitwiseAndTillN(n));
# This code is contributed by AnkitRai01
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the bitwise &
// of all the even numbers upto N
static int bitwiseAndTillN(int n)
{
// Initialize result as 2
int result = 2;
for (int i = 4; i <= n; i = i + 2)
{
result = result & i;
}
return result;
}
// Driver code
public static void Main()
{
int n = 2;
Console.WriteLine(bitwiseAndTillN(n));
}
}
// This code is contributed by AnkitRai01
<script>
// Javascript implementation of the above approach
// Function to return the bitwise &
// of all the even numbers upto N
function bitwiseAndTillN(n) {
// Initialize result as 2
let result = 2;
for (let i = 4; i <= n; i = i + 2) {
result = result & i;
}
return result;
}
// Driver code
let n = 2;
document.write(bitwiseAndTillN(n));
</script>
Time Complexity: O(n), where n is the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient approach: Efficient approach is to return 2 for N less than 4 and return 0 for all N>=4 because bitwise and of 2 and 4 is 0 and bitwise and of 0 with any number is 0.
Below is the implementation of the approach:
C++
// C++ implementation of the above approach
#include <iostream>
using namespace std;
// Function to return the bitwise &
// of all the numbers upto N
int bitwiseAndTillN(int n)
{
if (n < 4)
return 2;
else
return 0;
}
int main()
{
int n = 2;
cout << bitwiseAndTillN(n);
return 0;
}
// Java implementation of the above approach
class GFG
{
// Function to return the bitwise &
// of all the numbers upto N
static int bitwiseAndTillN(int n)
{
if (n < 4)
return 2;
else
return 0;
}
// Driver code
public static void main (String[] args)
{
int n = 2;
System.out.println(bitwiseAndTillN(n));
}
}
// This code is contributed by AnkitRai01
# Python3 implementation of the above approach
# Function to return the bitwise &
# of all the numbers upto N
def bitwiseAndTillN( n):
if (n < 4):
return 2
else:
return 0
# Driver code
n = 2
print(bitwiseAndTillN(n))
# This code is contributed by ANKITKUMAR34
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the bitwise &
// of all the numbers upto N
static int bitwiseAndTillN(int n)
{
if (n < 4)
return 2;
else
return 0;
}
// Driver code
public static void Main()
{
int n = 2;
Console.WriteLine(bitwiseAndTillN(n));
}
}
// This code is contributed by AnkitRai01
<script>
// JavaScript implementation of the above approach
// Function to return the bitwise &
// of all the numbers upto N
function bitwiseAndTillN(n)
{
if (n < 4)
return 2;
else
return 0;
}
// driver code
let n = 2;
document.write (bitwiseAndTillN(n));
// this code is contributed by shivanisinghss2110
</script>
Time complexity: O(1)
Auxiliary Space: O(1)
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