Check Even or Odd (original) (raw)

Last Updated : 9 Apr, 2026

Given a number **n, check whether it is even or odd. Return **true for even and **false for odd.

**Examples:

**Input: n = 15
**Output: false
**Explanation:15 % 2 = 1, so 15 is odd .

**Input: n = 44
**Output: true
**Explanation: 44 % 2 = 0, so 44 is even.

Try It Yourself

Table of Content

[Naive Approach] By Finding the Remainder - O(1) Time and O(1) Space
[Efficient Approach] Using Bitwise AND Operator - O(1) Time and O(1) Space

[Naive Approach] By Finding the Remainder - O(1) Time and O(1) Space

We can check the remainder when divided by 2. If the remainder is 0, the number is even, otherwise it is odd. For example, when we divide 13 by 2, we get remainder as 1 and when we divide 14 by 2, we get remainder as 0.

C++ `

#include using namespace std;

bool isEven(int n) { // finding remainder of n int rem = n % 2; if(rem == 0){ return true; } else{ return false; } }

int main() { int n = 15; if (isEven(n)) cout << "true"; else cout << "false";

return 0;

}

C

#include <stdio.h> #include <stdbool.h>

bool isEven(int n) { return n % 2 == 0; }

// Driver Code int main() { int n = 15;

if (isEven(n))
    printf("true\n");
else
    printf("false\n");

return 0;

}

Java

class GfG {

public static boolean isEven(int n)
{
    // finding remainder of n
    int rem = n % 2;
    if(rem == 0){
        return true;
    }
    else{
        return false;
    }
}

// Driver Code
public static void main(String[] args)
{
    int n = 15;
    if (isEven(n) == true)
        System.out.print("true");
    else
        System.out.print("false");
}

}

Python

def isEven(n):

# finding remainder of n
rem = n % 2; 
if rem == 0:
    return True
else:
    return False

if name == "main": n = 15 if isEven(n): print("true") else: print("false")

C#

using System; class Program { static bool isEven(int n) { // finding remainder of n int rem = n % 2; if (rem == 0){ return true; } else{ return false; } }

static void Main()
{
    int n = 15;
    if (isEven(n))
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

function isEven(n) {

// finding remainder of n
let rem = n % 2;
if(rem == 0){
    return true;
}
else{
    return false;
}

}

// Driver code let n = 15; if (isEven(n)) { console.log("true"); } else { console.log("false"); }

[Efficient Approach] Using Bitwise AND Operator - O(1) Time and O(1) Space

The last bit of all odd numbers is always 1, while for even numbers it’s 0. So, when performing bitwise AND operation with 1, odd numbers give 1, and even numbers give 0.

**Note: Bitwise operators are extremely **fast and efficient because they operate directly at the binary level, making them significantly faster than arithmetic or logical operations.

Examples:

15 -> 1 1 1 1
& 0 0 0 1
-------
0 0 0 1 , so this we can say it is an odd number.

44 -> 1 0 1 1 0 0
& 0 0 0 0 0 1
----------
0 0 0 0 0 0 , so this we can say it is an even number.

C++ `

#include using namespace std;

bool isEven(int n) {
// taking bitwise and of n with 1 if ((n & 1) == 0) return true; else return false; }

int main() { int n = 15; if (isEven(n) == true) cout << "true"; else cout << "false";

return 0;

}

C

#include <math.h> #include <stdio.h> #include <stdbool.h> // for using boolean datatype in c

bool isEven(int n) { // taking bitwise and of n with 1 if ((n & 1) == 0) return true; else return false; }

int main() { int n = 15; if (isEven(n)) { printf("true"); } else { printf("false"); } return 0; }

Java

class GfG { public static boolean isEven(int n) { // taking bitwise and of n with 1 if ((n & 1) == 0) return true; else return false; }

public static void main(String[] args)
{
    int n = 15;
    if (isEven(n) == true)
        System.out.print("true");
    else
        System.out.print("false");
}

}

Python

def isEven(n): # taking bitwise and of n with 1 if (n & 1) == 0: return True else: return False

if name == "main": n = 15 if isEven(n): print("true") else: print("false")

C#

using System;

class GfG { public static bool isEven(int n) { // taking bitwise and of n with 1 if ((n & 1) == 0) return true; else return false; }

public static void Main()
{
    int n = 15;
    if (isEven(n) == true)
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

function isEven(n) { // taking bitwise and of n with 1 if ((n & 1) === 0) { return true; } else { return false; } }

// Driver Code let n = 15; if (isEven(n)) { console.log("true"); } else { console.log("false"); }