CodeWars: One Line Task Check Range Solution

Approach

i < x == i > y:

• if both false, then condition will be true
• there will be no both true on both, so other will result in false

map here is just for iterating, the point here is accumulated c

array|c to grab c, because binary representation of array (NaN) OR a number will always result in that number

Implementation

1// prettier-ignore
2checkRange=(a,x,y,c=0)=>a.map(i=>c+=i<x==i>y)|c