Date within range?

Question

Given a MM/DD date (12/24) and a start/end date range (11/01 - 06/24), figure out if the date is within the listed date span.

Date ranges can be sequential (05/01 - 11/01) or wrap around to the next year (11/01 - 05/01).

Examples:

• 12/24 is in 11/01 - 06/24 = True
• 06/24 is in 11/01 - 06/24 = True
• 06/24 is in 06/24 - 06/24 = True
• 06/24 is in 11/01 - 06/23 = False
• 07/24 is in 11/01 - 06/24 = False
• 07/24 is in 05/01 - 11/01 = True
• 07/24 is in 07/23 - 07/20 = True

Years do not mater. The date and/or date range is assumed to apply for any year past or future.

The end date will always be after the start date. If the end date is numerically less than the start date, we assume we are wrapping around to the next year.

The three input variables can be via any method (args, HTTP, stdin, etc..) and in any format (string, JSON, array, etc..)

Response can be boolean or any form of a yes/no/correct/etc.. string. Shortest code wins.

You cannot simply pass args to a built-in function.

Answer 1

Python, 31 bytes

lambda a,d,e:(e<d)>=(a<d)+(e<a)

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Takes inputs as MM/DD in order target, start, end.

Answer 2

JavaScript (ES6), 26 20 bytes

(a,b,c)=>b>c^b>a^a>c

Output is 0 if a lies within the range b...c, 1 if not. Edit: Saved 6 bytes thanks to @nwellnhof.

Answer 3

Haskell, 28 bytes

(a%d)e=((e<a)/=(a<d))==(e<d)

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Thanks to Leo for the trick of xor'ing the Booleans with /=.

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Haskell, 33 bytes

(a%d)e=show[e>=d,d>a,a>e]!!16>'a'

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Answer 4

JavaScript (ES6), 31 bytes

Takes input as 3 strings in MM/DD format: date, range_from, range_to. Returns a boolean.

(a,b,c)=>!(b<c?a<b|a>c:a<b&a>c)

Test cases

let f =

(a,b,c)=>!(b<c?a<b|a>c:a<b&a>c)

console.log(f('12/24','11/01','06/24')) // True
console.log(f('06/24','11/01','06/24')) // True
console.log(f('06/24','06/24','06/24')) // True
console.log(f('06/24','11/01','06/23')) // False
console.log(f('07/24','11/01','06/24')) // False
console.log(f('07/24','05/01','11/01')) // True
console.log(f('07/24','07/23','07/20')) // True

Answer 5

Haskell, 38 37 bytes

(b#e)d=(last$and:[or|b>e])[d>=b,d<=e]

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Saved 1 byte thanks to Laikoni

Answer 6

Perl 6, 33 31 29 bytes

{.[0]>.[2]??![>] $_!![<=] $_}

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Takes a list containing start, date, and end.

-2 bytes thanks to Ramillies.

Answer 7

Perl 5, 60 bytes

59 bytes of code + 1 -a

$F[1]=~s/./2/if$F[1]le$F[0];say($F[2]le$F[1]&&$F[2]ge$F[0])

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Input: start_date end_date target_date