Print “Even” or “Odd” without using conditional statement

Write a program that accepts a number from the user and prints “Even” if the entered number is even and prints “Odd” if the number is odd. You are not allowed to use any comparison (==, <,>,…etc) or conditional statements (if, else, switch, ternary operator,. Etc).

Method 1
Below is a tricky code can be used to print “Even” or “Odd” accordingly. 

C++

#include <iostream>

using namespace std;

int main()
{
    char arr[2][5] = { "Even", "Odd" };
    int no;
    cout << "Enter a number: ";
    cin >> no;
    cout << arr[no % 2];
    getchar();
    return 0;
}

Java

import java.util.Scanner;
class GFG
{
    public static void main(String[] args)
    {

        String[] arr = {"Even", "Odd"};

        Scanner s = new Scanner(System.in);

        System.out.print("Enter the number: ");
        int no = s.nextInt();

        System.out.println(arr[no%2]);
    }
}

// This code is contributed by divyeshrabadiya07.

Python3

arr = ["Even", "Odd"]
print ("Enter the number")
no = int(input())
print (arr[no % 2])

C#

using System;
class GFG {
  static void Main() {
    string[] arr = {"Even", "Odd"};

    Console.Write("Enter the number: ");

    string val;
    val = Console.ReadLine();
    int no = Convert.ToInt32(val);

    Console.WriteLine(arr[no%2]);
  }
}

// This code is contributed by divyesh072019.

PHP

<?php
$arr = ["Even", "Odd"];
$input = 5;
echo ($arr[$input % 2]);

// This code is contributed
// by Aman ojha
?>

Javascript

<script>

    let arr = ["Even", "Odd"];
    let no = prompt("Enter a number: ");

    document.write(arr[no % 2]);

   // This code is contributed by suresh07

</script>

Time Complexity: O(1)
Auxiliary Space: O(1)

Method 2
Below is another tricky code can be used to print “Even” or “Odd” accordingly. Thanks to student for suggesting this method.

C++

#include <iostream>
using namespace std;

int main()
{
    int no = 8;

    (no & 1 && cout << "odd" )|| cout << "even";

    return 0;
}

// This code is contributed by sarajadhav12052009

C

#include<stdio.h>
int main()
{
    int no = 8;
    (no & 1 && printf("odd"))|| printf("even");
    return 0;
}

Java

import java.util.*;

class GFG
{
public
    static void main(String[] args)
    {
        int no = 8;

        if ((no & 1) != 0)
        {
            System.out.println("odd");
        }
        else
        {
            System.out.println("even");
        }
    }
}

Python3

no = 8
if no & 1:
    print("odd")
else:
    print("even")

C#

using System;

class GFG
{
    static void Main(string[] args)
    {
        int no = 8;

        if ((no & 1) != 0)
        {
            Console.WriteLine("odd");
        }
        else
        {
            Console.WriteLine("even");
        }
    }
}

Javascript

let no = 8;

(no & 1 && console.log("odd")) || console.log("even");

Output

even

Time Complexity: O(1)
Auxiliary Space: O(1)

Method 3
This can also be done using a concept known as Branchless Programming. Essentially, make use of the fact that a true statement in Python (other some other languages) evaluates to 1 and a false statement evaluates to false.

C++

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cout << "Enter a number: ";
    cin >> n;

    if (n % 2 == 0) {
        cout << "Even" << endl;
    }
    else {
        cout << "Odd" << endl;
    }

    return 0;
}

// This code is contributed by surajrasr7277.

Java

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
          int n = 8;
        if (n % 2 == 0) {
            System.out.println("Even");
        }
        else {
            System.out.println("Odd");
        }
    }
}

Python3

# code
n = 8
print("Even" * (n % 2 == 0), "Odd" * (n % 2 != 0))

Javascript

let n = 8;

if (n%2==0)
    document.write("Even");
else
    document.write("Odd");

C#

// This code checks whether a given number is even or odd.
using System;

class MainClass {
    public static void Main()
    {
        Console.Write("Enter a number: ");
        int n = Convert.ToInt32(Console.ReadLine());
        if (n % 2 == 0) {
            Console.WriteLine("Even");
        }
        else {
            Console.WriteLine("Odd");
        }
    }
}

Output

Enter a number: Even 

Time Complexity: O(1)
Auxiliary Space: O(1)

Please write comments if you find the above code incorrect, or find better ways to solve the same problem