Check whether a given number is even or odd
Given a number, check whether it is even or odd.
Examples :
Input: 2 Output: even Input: 5 Output: odd
One simple solution is to find the remainder after dividing by 2.
C++
// A simple C++ program to // check for even or odd #include <iostream> using namespace std; // Returns true if n is // even, else odd bool isEven( int n) { return (n % 2 == 0); } // Driver code int main() { int n = 101; isEven(n) ? cout << "Even" : cout << "Odd" ; return 0; } |
Java
// Java program to // check for even or odd class GFG { // Returns true if n is even, else odd public static boolean isEven( int n) { return (n % 2 == 0 ); } // Driver code public static void main(String[] args) { int n = 101 ; if (isEven(n) == true ) System.out.print( "Even" ); else System.out.print( "Odd" ); } } // This code is contributed by rishabh_jain |
Python3
# A simple Python3 code # to check for even or odd # Returns true if n is even, else odd def isEven(n): return (n % 2 = = 0 ) # Driver code n = 101 print ( "Even" if isEven(n) else "Odd" ) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to // check for even or odd using System; class GFG { // Returns true if n is even, else odd public static bool isEven( int n) { return (n % 2 == 0); } // Driver code public static void Main() { int n = 101; if (isEven(n) == true ) Console.WriteLine( "Even" ); else Console.WriteLine( "Odd" ); } } // This code is contributed by vt_m |
PHP
<?php // A simple PHP program to // check for even or odd // Returns true if n is // even, else odd function isEven( $n ) { return ( $n % 2 == 0); } // Driver code $n = 101; if (isEven != true) echo "Even" ; else echo "Odd" ; // This code is contributed by Ajit ?> |
Javascript
<script> // A simple Javascript program to // check for even or odd // Returns true if n is // even, else odd function isEven(n) { return (n % 2 == 0); } // Driver code let n = 101; isEven(n) ? document.write( "Even" ) :document.write( "Odd" ); // This code is contributed by Mayank Tyagi </script> |
Time Complexity: O(1)
Auxiliary Space: O(1)
A better solution is to use bitwise operators. We need to check whether last bit is 1 or not. If last bit is 1 then number is odd, otherwise always even.
Explanation:
input : 5 // odd 00000101 & 00000001 -------------- 00000001 -------------- input : 8 //even 00001000 & 00000001 -------------- 00000000 --------------
Below is the implementation of the idea.
C++
// A simple C++ program to // check for even or odd #include <iostream> using namespace std; // Returns true if n is // even, else odd bool isEven( int n) { // n & 1 is 1, then // odd, else even return (!(n & 1)); } // Driver code int main() { int n = 101; isEven(n)? cout << "Even" : cout << "Odd" ; return 0; } |
C
#include <stdio.h> #include <math.h> int main(){ int n = 101; if (n%2==0){ printf ( "Even" ); } else { printf ( "Odd" ); } return 0; } |
Java
// Java program to // check for even or odd class GFG { // Returns true if n // is even, else odd public static boolean isEven( int n) { if ((n & 1 ) == 0 ) return true ; else return false ; } // Driver code public static void main(String[] args) { int n = 101 ; if (isEven(n) == true ) System.out.print( "Even" ); else System.out.print( "Odd" ); } } // This code is contributed by rishabh_jain |
Python3
# A Python3 code program # to check for even or odd # Returns true if n is even, else odd def isEven(n): # n&1 is 1, then odd, else even return ( not (n & 1 )) # Driver code n = 101 ; print ( "Even" if isEven(n) else "Odd" ) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to // check for even or odd using System; class GFG { // Returns true if n // is even, else odd public static bool isEven( int n) { if ((n & 1) == 0) return true ; else return false ; } // Driver code public static void Main() { int n = 101; if (isEven(n) == true ) Console.WriteLine( "Even" ); else Console.WriteLine( "Odd" ); } } // This code is contributed by vt_m. |
PHP
<?php // A simple PHP program to // check for even or odd // Returns true if n is // even, else odd function isEven( $n ) { return (!( $n & 1)); } // Driver code $n = 101; if (isEven( $n ) == true) echo "Even" ; else echo "Odd" ; // This code is contributed by Smitha ?> |
Javascript
<script> // A simple JavaScript program to // check for even or odd // Returns true if n is // even, else odd function isEven(n) { // n & 1 is 1, then // odd, else even return (!(n & 1)); } // Driver code let n = 101; isEven(n)? document.write( "Even" ) : document.write( "Odd" ); // This code is contributed by Manoj. </script> |
Odd
Time Complexity: O(1)
Auxiliary Space: O(1)
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