A number that is divisible by 2 and generates a remainder of 0 is called an even number. All the numbers ending with 0, 2, 4, 6, and 8 are even numbers. On the other hand, number that is not divisible by 2 and generates a remainder of 1 is called an odd number. All the numbers ending with 1, 3, 5,7, and 9 are odd numbers. Do refer to the below illustration to get what is supposed to be conveyed out basics here via generic Illustration for any random integer, check whether it is even or odd.
Input : 13 Output: ODD
Input : 24 Output: EVEN
Methods:
There are various ways to check whether the given number is odd or even. Some of them are as follows starting from the brute force approach ending up at the most optimal approach.
- Using Brute Force- Naive Approach
- Using bitwise operators
- Using Bitwise OR
- Using Bitwise AND
- Using Bitwise XOR
- By Checking the Least Significant Bit
Method 1: Brute Force Naive Approach
It is to check the remainder after dividing by 2. Numbers that are divisible by 2 are even else odd.
Example
Java
// Java Program to Check if Given Integer is Odd or Even// Using Brute Forcew Approach// Importing required classesimport java.io.*;import java.util.Scanner;// Main classclass GFG { // Main Driver Method public static void main(String[] args) { // Declaring and initializing integer variable int num = 10; // Checking if number is even or odd number // via remainder if (num % 2 == 0) { // If remainder is zero then this number is even System.out.println("Entered Number is Even"); } else { // If remainder is not zero then this number is // odd System.out.println("Entered Number is Odd"); } }} |
Entered Number is Even
Time Complexity: O(1)
Auxiliary Space: O(1)
Now let us dwell on optimal approaches as follows below as follows with help of bitwise operators
Method 2: Using bitwise operators
- Bitwise OR
- Bitwise AND or Bitwise XOR
2-A: Using Bitwise OR
Bitwise OR operation of the even number by 1 increment the value of the number by 1 otherwise it remains unchanged.
Illustration: Bitwise OR
Number = 12 1 1 0 0 - Representation of 12 in Binary Format
Bitwise OR 0 0 0 1 - Representation of 1 in Binary Format
1 1 0 1 - Representation of 13 in Binary Format
Result- Number was even so bitwise Or by 1 increment the value by 1
Number = 15 1 1 1 1 - Representation of 15 in Binary Format
Bitwise OR 0 0 0 1 - Representation of 1 in Binary Format
1 1 1 1 - Representation of 15 in Binary Format
Result- Number was odd so bitwise Or by 1 doesn't increment the value by 1
Example
Java
// Java Program to Check if Given Integer is Odd or Even// Using Bitwise OR// Importing required classesimport java.util.*;// Main classpublic class GFG { // Main driver method public static void main(String[] args) { // Declaring and initializing integer variable // to be checked int n = 100; // Condition check // if n|1 if greater than n then this number is even if ((n | 1) > n) { // Print statement System.out.println("Number is Even"); } else { // Print statement System.out.println("Number is Odd"); } }} |
Number is Even
Time Complexity: O(1)
Auxiliary Space: O(1)
2-B: Using Bitwise AND
Bitwise AND operation of the odd number by 1 will be 1 because the last bit will be already set otherwise it will give 0.
Illustration: Bitwise AND
Number = 5 0 1 0 1 - Representation of 5 in Binary Format
Bitwise AND 0 0 0 1 - Representation of 1 in Binary Format
0 0 0 1 - Representation of 1 in Binary Format
Result- Number was odd so bitwise And by 1 is 1
Number = 8 1 0 0 0 - Representation of 8 in Binary Format
Bitwise AND 0 0 0 1 - Representation of 1 in Binary Format
0 0 0 0 - Representation of 0 in Binary Format
Result- Number was even so bitwise And by 1 is 0
Example
Java
// Java Program to Check if Given Integer is Odd or Even// Using Bitwise AND// Importing required classesimport java.util.*;// Main classpublic class GFG { // Main driver method public static void main(String[] args) { // Declare and initializing integer variable int n = 91; // Condition Check // Bitwise AND of any odd number by 1 gives 1 if ((n & 1) == 1) { // Print statement System.out.println("Number is Odd"); } else { // Print statement System.out.println("Number is Even"); } }} |
Number is Odd
Time Complexity: O(1)
Auxiliary Space: O(1)
2-C: Using Bitwise XOR
Bitwise XOR operation of the even number by 1 increment the value of the number by 1 otherwise it decrements the value of the number by 1 if the value is odd. It is the most optimal approach.
Illustration: Bitwise XOR
Number = 5 0 1 0 1 - Representation of 5 in Binary Format
Bitwise XOR 0 0 0 1 - Representation of 1 in Binary Format
0 1 0 0 - Representation of 4 in Binary Format
Result- Number was odd so bitwise And by 1 decrement the value
Number = 8 1 0 0 0 - Representation of 8 in Binary Format
Bitwise XOR 0 0 0 1 - Representation of 1 in Binary Format
1 0 0 1 - Representation of 9 in Binary Format
Result- Number was even so bitwise And by 1 increment the value
Example
Java
// Java Program to Check if Given Integer is Odd or Even// Using Bitwise XOR// Importing required classesimport java.util.*;// Main classpublic class GFG { // Main driver method public static void main(String[] args) { // Declare and initializing integer variable int num = 99; // Condition Check // if number^1 increments by 1 then its even number, // else odd if ((num ^ 1) == num + 1) { // Print statement System.out.println("Number is Even"); } else { // Print statement System.out.println("Number is Odd"); } }} |
Number is Odd
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 3: Checking the LSB of the Number
The LSB(Least Significant Bit) of an even number is always 0 and that of an odd number is always 1.
Example
Java
// Java Program to Check if Given Integer is Odd or Even// by checking the LSB of the Number// Importing required classesimport java.util.*;// Main class// TestEvenOddByCheckingLSBpublic class GFG { // Method 1 // To test number is even or odd public static String testOddEvenByCheckingLSB(int a) { if (a != 0) { if (Integer.toBinaryString(a).endsWith("0")) { return "Even"; } else { return "Odd"; } } // Here we will land if // it does not ends with 0 else { return "Zero"; } } // Method 2 // Main driver method public static void main(String[] args) { // Iterationg over using for loop for (int i = 0; i <= 10; i++) { // Calling the function and printing // corresponding number is even or odd System.out.println( i + " : " + testOddEvenByCheckingLSB(i)); } }} |
0 : Zero 1 : Odd 2 : Even 3 : Odd 4 : Even 5 : Odd 6 : Even 7 : Odd 8 : Even 9 : Odd 10 : Even
Time Complexity: O(1)
Auxiliary Space: O(1)
