Check if a number is even or odd without using loop or if- else #java #program #java4quicklearning

Check if a number is even or odd without using loop or if- else #java #program #java4quicklearning

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[nvclas]

number & 1 == 0 -> Even
number & 1 == 1 -> Uneven

The absolute best way to check is bitwise operation

[zacjohnson]

For those who don't code:

N = input number to check if even/odd
res(0) = the first option in list, in this case even
res(1) = the second option, in this case odd
% = returns the remainder of A÷B when A%B

The code says to divide N by 2 and return the remainder, this checks whether it is odd or even because any even number will have a remainder of 0 after being divided by 2, while the opposite is true for odd numbers (they will always have a remainder of 1). Then you put the remainder into the res list and if the remainder is 1(odd), res(1) = odd and if the remainder is 0(even), res(0) = even.

[mvidyakrishna]

Dayum
Solving problems makes one humble.

[TheAkshaysWay]

In next 1000 years I won't find this way of even or odd

[shalevforfor]

you can also use n & 1 it will work as well and be more

[SteveTheNerd]

If you’re write a function returning a Boolean, literally no if statement is required. Return n % 2 == 0

[chupapimunanyo]

Return ((N/2)+““).split(“.“).length == 2

[LuvRatan-xs]

print(n%2==0?"Even":"odd")

[names-mars]

He’s not a genius, simply resourceful. The correct way to do it is bit wise operation or ternary operation if the language supports it.

[kartik-R]

num = 11
dict = {0:"Even", 1:"Odd"}
print(dict[num%2])

[sujalsinha]

number = 12
print({0: "Even", 1: "Odd"}[number % 2]) # Output: Even

[Sourav_gupta_]

C program
Int num;
Printf("enter");
Scanf("%d", &num);
Num%2==0 ? printf("even" ) : printf("odd");🎉

[bogdogun]

Python:
print((n%2) * 'odd' + (1-n%2) * 'even')

[pavansharma]

Output is:Odd
Because n%2 = 17%2 =1;
On Index 1 is "Odd"

[neozetterberg]

That’s modulus. Yes, it is very fun and interesting.

[takatoo]

but modules are allowed. just import OddEven bro 😊

[minhkhoi]

python
n = 17
print("Even" * (1 - (n % 2)) + "Odd" * (n % 2))

[aditya]

Predicate<Integer> isEven= i -> i%2==0 ;

Sop (isEven.test(4)); : true
Sop(isEven.test(5)); : false

[NilsCoussement-]

You could switch out the res array with translations. Genious!

[staenker]

A True coder would say building a string is not checking… the checking is done by the human comparing the string in that case. People who would give flawed and imprecise tasks…