challenge - #3 if statements from hackerrank

#condition

1. N == odd , print out ‘weird’

2. n == even 2≤ n 5≤ , print out ‘not weird’

3. n==even and 6≤ n ≤20 , print ‘weird’

4. n==even and n >20 , print ‘not weird’

first of all, i need to figure out how to get an odd number.

isOdd = number % 2 
if isOdd = 0, the given number is even, otherwise the odd 
The remainder operator returns the remainder left over when one operand is divided by a second operand. It always takes the sign of the dividend.

once i figured it out, i’ve wrote like below.

i’m a little bit ashamed of mine.

function printNum(n){
    var num = Math.abs(n % 2);
    
    if (num == 1){
        console.log('weird')
    }else if( 2 <= n && n <= 5 ){
        console.log('not weird')
    }else if (6 <= n && n <= 20){
        console.log('weird')
    }else{
        console.log('not weird')
    }

then I found out other’s codes that was simplified .

function isOdd(n) {
    return n % 2 == 1;
}
function between(low, n, high) {
    return low <= n && n <= high;
}
function main11() {
    var n = parseInt(readLine());
console.log(isOdd(n) || between(6, n, 20) ? "Weird" : "Not Weird");
}
//or
function printNum(n){
    
    if ( n%2 == 1 || 6 <= n && n <= 20 ){
        console.log('weird')
    }else {
        console.log('not weird')
    }
}

concise code!

it’s hard for me to simplify and optimize the code.

간결한코드…쉽지않다.

홀수 or 6≤ n ≤20, print ‘weird’

#reference

https://www.hackerrank.com/challenges/30-conditional-statements/problem

https://jsfiddle.net/blossom/961fknh7/