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# Daily Challenge #277 - Over the Road

You've just moved into a perfectly straight street with exactly n identical houses on either side of the road. Naturally, you would like to find out the house number of the people on the other side of the street. The street looks something like this:

``````1|   |6
3|   |4
5|   |2
``````

Evens increase on the right; odds decrease on the left. House numbers start at 1 and increase without gaps. When n = 3, 1 is opposite 6, 3 opposite 4, and 5 opposite 2.

Given your house number address and length of street n, give the house number on the opposite side of the street.

### Examples

`over_the_road(1, 3)` = 6
`over_the_road(3, 3)` = 4

### Tests

`over_the_road(3, 5)`
`over_the_road(7, 11)`

Good luck!

This challenge comes from rge123 on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

## Discussion (5)

peter279k

Simple solution:

``````<?php
return \$street*2 - \$address + 1;
}
``````

Here is sample picture to explain all of this:

Bubbler • Edited on

APL (using Dyalog APL):

``````   ⍝ left(⍺):address, right(⍵):n
6
4
8
16

⍝ Short, "tacit" form
⍝ Arithmetic automatically maps over multiple numbers!
1 3 3 7 OverTheRoad 3 3 5 11
6 4 8 16
``````

(Use the bookmarklet on this post to see the code with APL font and syntax highlighting.)

Explanation: The sum of two houses facing each other is always the same for given `n`, which is `2n+1`. So the opposite house number becomes `2n+1-address`.

Rafi

Ruby

``````def over_the_road(address, n)
(n * 2) + 1 - address
end
``````
SavagePixie
``````const overTheRoad = (a, n) => 2 * n - a + 1
``````
Michael Kohl

Postscript:

``````GS>/overTheRoad { 2 mul exch sub 1 add } def