# 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 number## Implementation

1// prettier-ignore2checkRange=(a,x,y,c=0)=>a.map(i=>c+=i<x==i>y)|c

## Comments

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## Tags

codewars

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