/*
Description:
Scenario
Now that the competition gets tough it will Sort out the men from the boys .

Men are the Even numbers and Boys are the odd  !alt !alt

Task
Given an array/list [] of n integers , Separate The even numbers from the odds , or Separate the men from the boys  !alt !alt

Notes
Return an array/list where Even numbers come first then odds

Since , Men are stronger than Boys , Then Even numbers in ascending order While odds in descending .

Array/list size is at least 4 .

Array/list numbers could be a mixture of positives , negatives .

Have no fear , It is guaranteed that no Zeroes will exists . !alt

Repetition of numbers in the array/list could occur , So (duplications are not included when separating).

Input >> Output Examples:
1- menFromBoys ({7, 3 , 14 , 17}) ==> return ({14, 17, 7, 3})
Explanation:
Since , { 14 } is the even number here , So it came first , then the odds in descending order {17 , 7 , 3} .

2- menFromBoys ({-94, -99 , -100 , -99 , -96 , -99 }) ==> return ({-100 , -96 , -94 , -99})
Explanation:
Since , { -100, -96 , -94 } is the even numbers here , So it came first in ascending order , then the odds in descending order { -99 }

Since , (Duplications are not included when separating) , then you can see only one (-99) was appeared in the final array/list .

3- menFromBoys ({49 , 818 , -282 , 900 , 928 , 281 , -282 , -1 }) ==> return ({-282 , 818 , 900 , 928 , 281 , 49 , -1})
Explanation:
Since , {-282 , 818 , 900 , 928 } is the even numbers here , So it came first in ascending order , then the odds in descending order { 281 , 49 , -1 }

Since , (Duplications are not included when separating) , then you can see only one (-282) was appeared in the final array/list .
*/
function menFromBoys(arr) {
  return [
    ...new Set(
      arr
        .filter(v => v % 2 === 0)
        .sort((a, b) => a - b)
        .concat(arr.filter(v => v % 2 !== 0).sort((a, b) => b - a))
    )
  ];
}
