**Problem Statement**: Given a number n, check whether a given number is even or odd.

**Examples:**

Example 1:Input:n=5Output:oddExplanation:5 is not divisible by 2.Example 2:Input:n=6Output:evenExplanation:6 is divisible by 2.

**Solution**

*Disclaimer*: *Don’t jump directly to the solution, try it out yourself first.*

**Solution1: Using division**

**Intuition: **If we can prove that the given number is divisible by 2 then it will be even, otherwise it will be odd.

**Approach: **

- Find the remainder of the number.
- If it is 0 then the number is even otherwise it is odd.

**Code:**

## C++ Code

```
#include<bits/stdc++.h>
using namespace std;
int remainder(int n) {
return (n % 2);
}
int main() {
int n = 5;
if (remainder(n) == 0) {
cout<<n <<" is even.";
} else {
cout<<n<<" is odd.";
}
}
```

**Output:**

5 is odd.

**Time complexity:** O(1)

**Space Complexity:** O(1)

## Java Code

```
import java.util.*;
public class tuf {
public static void main(String[] args) {
int n = 5;
if (remainder(n) == 0) {
System.out.println(n + " is even.");
} else {
System.out.println(n + " is odd.");
}
}
public static int remainder(int n) {
return (n % 2);
}
}
```

**Output:**

5 is odd.

**Time complexity:** O(1)

**Space Complexity:** O(1)

**Solution2: Using the bitwise operator**

**Intuition: **As we know that bitwise operations are generally faster than normal operations so we should prefer it over normal operations.

Even Number: The last bit of even number is always 0.

Odd Number: Last bit of the odd number is always 1.

If we can get the last bit of any number then we can say that it is even or odd.

**Approach: **

- Take AND of n with 1.
- If it is 0 then print even otherwise odd.

**Code:**

## C++ Code

```
#include<bits/stdc++.h>
using namespace std;
int solve(int n) {
if ((n & 1) == 0)
return 0;
else
return 1;
}
int main() {
int n = 5;
if (solve(n) == 0) {
cout<<n<<" is even.";
} else {
cout<<n << " is odd.";
}
}
```

**Output:**

5 is odd.

**Time complexity:** O(1)

**Space Complexity:** O(1)

## Java Code

```
import java.util.*;
public class tuf {
public static void main(String[] args) {
int n = 5;
if (solve(n) == 0) {
System.out.println(n + " is even.");
} else {
System.out.println(n + " is odd.");
}
}
public static int solve(int n) {
if ((n & 1) == 0)
return 0;
else
return 1;
}
}
```

**Output:**

5 is odd.

**Time complexity:** O(1)

**Space Complexity:** O(1)

Special thanks toplease check out this articlePrashant Sahufor contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam,