Given a number N, the task is to check whether the number is even or odd using Bitwise Operators.
Examples:
Input: N = 11
Output: OddInput: N = 10
Output: Even
Following Bitwise Operators can be used to check if a number is odd or even:
1. Using Bitwise XOR operator:
The idea is to check whether the last bit of the number is set or not. If the last bit is set then the number is odd, otherwise even.
As we know bitwise XOR Operation of the Number by 1 increment the value of the number by 1 if the number is even otherwise it decrements the value of the number by 1 if the value is odd.
Below is the implementation of the above approach:
C++
// C++ program to check for even or odd// using Bitwise XOR operator#include <iostream>using namespace std;// Returns true if n is even, else oddbool isEven(int n){ // n^1 is n+1, then even, else odd if ((n ^ 1) == (n + 1)) return true; else return false;}// Driver codeint main(){ int n = 100; isEven(n) ? cout << "Even" : cout << "Odd"; return 0;} |
Java
// Java program to check for even or odd// using Bitwise XOR operatorclass GFG { // Returns true if n is even, else odd static boolean isEven(int n) { // n^1 is n+1, then even, else odd if ((n ^ 1) == (n + 1)) return true; else return false; } // Driver code public static void main(String[] args) { int n = 100; System.out.print(isEven(n) == true ? "Even" : "Odd"); }}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to check for even or odd# using Bitwise XOR operator# Returns true if n is even, else odddef isEven(n): # n^1 is n+1, then even, else odd if ((n ^ 1) == (n + 1)): return True else: return False# Driver codeif __name__ == "__main__": n = 100 print("Even") if isEven(n) else print("Odd")# This code is contributed by AnkitRai01 |
C#
// C# program to check for even or odd// using Bitwise XOR operatorusing System;class GFG { // Returns true if n is even, else odd static bool isEven(int n) { // n^1 is n+1, then even, else odd if ((n ^ 1) == (n + 1)) return true; else return false; } // Driver code public static void Main(String[] args) { int n = 100; Console.Write(isEven(n) == true ? "Even" : "Odd"); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program to check for even or odd// using Bitwise XOR operator// Returns true if n is even, else oddfunction isEven(n){ // n^1 is n+1, then even, else odd if ((n ^ 1) == (n + 1)) return true; else return false;}// Driver code let n = 100; isEven(n) ? document.write("Even"): document.write("Odd");// This code is contributed by Surbhi Tyagi.</script> |
Even
Time Complexity: O(1)
Auxiliary Space: O(1)
2. Using Bitwise AND operator:
The idea is to check whether the last bit of the number is set or not. If last bit is set then the number is odd, otherwise even.
As we know bitwise AND Operation of the Number by 1 will be 1, If it is odd because the last bit will be already set. Otherwise, it will give 0 as output.
Below is the implementation of the above approach:
C++
// C++ program to check for even or odd// using Bitwise AND operator#include <iostream>using namespace std;// Returns true if n is even, else oddbool isEven(int n){ // n&1 is 1, then odd, else even return (!(n & 1));}// Driver codeint main(){ int n = 101; isEven(n) ? cout << "Even" : cout << "Odd"; return 0;} |
Java
// Java program to check for even or odd// using Bitwise AND operatorclass GFG { // Returns true if n is even, else odd static boolean isEven(int n) { // n&1 is 1, then odd, else even return ((n & 1) != 1); } // Driver code public static void main(String[] args) { int n = 101; System.out.print(isEven(n) == true ? "Even" : "Odd"); }}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to check for even or odd# using Bitwise AND operator# Returns true if n is even, else odddef isEven(n): # n&1 is 1, then odd, else even return (not(n & 1))# Driver codeif __name__ == "__main__": n = 101 if isEven(n): print("Even") else: print("Odd")# This code is contributed by AnkitRai01 |
C#
// C# program to check for even or odd// using Bitwise AND operatorusing System;class GFG { // Returns true if n is even, else odd static bool isEven(int n) { // n&1 is 1, then odd, else even return ((n & 1) != 1); } // Driver code public static void Main() { int n = 101; Console.Write(isEven(n) == true ? "Even" : "Odd"); }}// This code is contributed by AnkitRai01 |
Javascript
<script>// javascript program to check for even or odd// using Bitwise AND operator // Returns true if n is even, else odd function isEven(n) { // n&1 is 1, then odd, else even return ((n & 1)!=1); } // Driver code var n = 101; document.write(isEven(n) == true ? "Even" : "Odd"); // This code is contributed by Princi Singh </script> |
Odd
Time Complexity: O(1)
Auxiliary Space: O(1)
3. Using Bitwise OR operator: The idea is to check whether the last bit of the number is set or not. If the last bit is set then the number is odd, otherwise even. As we know bitwise OR Operation of the Number by 1 increment the value of the number by 1 if the number is even otherwise it will remain unchanged. So, if after OR operation of number with 1 gives a result that is greater than the number then it is even and we will return true otherwise it is odd and we will return false.
Below is the implementation of the above approach:
C++
// C++ program to check for even or odd// using Bitwise OR operator#include <iostream>using namespace std;// Returns true if n is even, else oddbool isEven(int n){ // n|1 is greater than n, then even, else odd if ((n | 1) > n) return true; else return false;}// Driver codeint main(){ int n = 100; isEven(n) ? cout << "Even" : cout << "Odd"; return 0;} |
Java
// Java program to check for even or odd// using Bitwise OR operatorclass GFG { // Returns true if n is even, else odd static boolean isEven(int n) { // n|1 is greater than n, then even, else odd if ((n | 1) > n) return true; else return false; } // Driver code public static void main(String[] args) { int n = 100; System.out.print(isEven(n) == true ? "Even" : "Odd"); }} |
Python3
# Python3 program to check for even or odd# using Bitwise OR operator# Returns true if n is even, else odddef isEven(n): # n|1 is greater than n, then even, else odd if (n | 1 > n): return True else: return False# Driver codeif __name__ == "__main__": n = 100 print("Even") if isEven(n) else print("Odd") |
C#
// C# program to check for even or odd// using Bitwise XOR operatorusing System;class GFG { // Returns true if n is even, else odd static bool isEven(int n) { // n|1 is greater than n, then even, else odd if ((n | 1) > n) return true; else return false; } // Driver code public static void Main(String[] args) { int n = 100; Console.Write(isEven(n) == true ? "Even" : "Odd"); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// javascript program to check for even or odd // using Bitwise OR operator // Returns true if n is even, else odd function isEven(n) { // n|1 is greater than n, then even, else odd if ((n | 1) > n) return true; else return false; } // Driver code var n = 100; document.write(isEven(n) == true ? "Even" : "Odd"); // This code is contributed by Amit Katiyar. </script> |
Even
Time Complexity: O(1)
Auxiliary Space: O(1)
4. Using bitwise left and right shift operators: The idea is to check whether a number remains the same after performing some operations i.e. bitwise left and right shift. When we do a bitwise right shift of the number then the last bit of the number is removed whenever it is 1 or 0. After that performing the left shift to the number by default a new binary is added to the last bit which is 0, by displacement of one bit by one. Even numbers remain the same but odd changed its value. compare the initial and final values of the number to check number is even or odd.
Below is the implementation of the above approach.
C++
#include <iostream>using namespace std;int main(){ int num = 2; // To take input from user, prefer below code // int num; // cout << "Enter a number: \n"; // cin >> num; if (num == (num >> 1) << 1) { cout << num << " is even." << endl; } else { cout << num << " is not even." << endl; } // Below is short hand // num == (num >> 1) << 1 ? cout << num << " is even." // << endl : cout << num << " is not even." << endl; return 0;}// This code is contributed by lokeshpotta20 |
C
#include <stdio.h>int main() { int num = 2 ; // To take input from user, prefer below code // int num ; // printf("Enter a number: \n"); // scanf("%d",&num); if(num == ( num>>1 ) << 1){ printf("%d is even.\n",num); } else{ printf("%d is not even.\n",num); } // below is short hand // num==(num>>1)<<1? printf("%d is even.\n",num):printf("%d is odd.\n",num); return 0;} |
Java
// Java program to check for even or odd// using Bitwise OR operatorimport java.util.*;class GFG { // Driver code public static void main(String[] args) { int num = 2 ; // To take input from user, prefer below code // int num ; // System.out.println("Enter a number: \n"); // num = in.nextInt(); if(num == ( num>>1 ) << 1){ System.out.println(num + " is even.\n"); } else{ System.out.println(num + " is not even.\n"); } // below is short hand // num==(num>>1)<<1? System.out.println(num + " is even.\n"):System.out.println(num + "%d is not even.\n"); }}// This code is contributed by abhijeet19430. |
Python3
# Python3 program to check for even or odd# using Bitwise OR operatornum = 2 ;# To take input from user, prefer below code# int num ;# printf("Enter a number: \n");# scanf("%d",&num);if(num == ( num>>1 ) << 1): print(num, "is even.\n");else: print(num, "is not even.\n");# below is short hand # num==(num>>1)<<1? printf("%d is even.\n",num):printf("%d is odd.\n",num); |
C#
using System;using System.Collections.Generic;class GFG { public static void Main() { int num = 2 ; // To take input from user, prefer below code // int num ; // printf("Enter a number: \n"); // scanf("%d",&num); if(num == ( num>>1 ) << 1){ Console.Write(num+" is even.\n"); } else{ Console.Write(num+" is not even.\n"); } // below is short hand // num==(num>>1)<<1? printf("%d is even.\n",num):printf("%d is odd.\n",num); }}// This code is contributed by ratiagarwal. |
Javascript
let num = 2 ;// To take input from user, prefer below code// int num ;// printf("Enter a number: \n");// scanf("%d",&num);if(num == ( num>>1 ) << 1){ console.log(num, "is even.\n"); }else{ console.log(num, "is not even.\n"); }// below is short hand // num==(num>>1)<<1? printf("%d is even.\n",num):printf("%d is odd.\n",num);// This code is contributed by poojaagarwal2. |
2 is even.
Time Complexity: O(1)
Auxiliary Space: O(1)
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