CodeWars: One Line Task Check Range Solution
Approach
i < x == i > y
:- if both false, then condition will betrue
- there will be no both trueon both, so other will result infalse
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 numberImplementation
1// prettier-ignore2checkRange=(a,x,y,c=0)=>a.map(i=>c+=i<x==i>y)|c
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