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NOTE: Since I'm Dutch myself, all dates are in the Dutch dd-MM-yyyy format in the challenge description and test cases.

Challenge:

Inputs:
Start date \$s\$; End date \$e\$; Digit \$n\$

Outputs:
All dates within the range \$[s,e]\$ (including on both sides), which contain \$n\$ amount of unique digits in their date.

Example:

Inputs: Start date: 12-11-1991; End date: 02-02-1992; Digit: 4

Outputs:
With leading 0s for days/months:

[20-11-1991, 23-11-1991, 24-11-1991, 25-11-1991, 26-11-1991, 27-11-1991, 28-11-1991, 30-11-1991, 01-12-1991, 02-12-1991, 09-12-1991, 10-12-1991, 13-12-1991, 14-12-1991, 15-12-1991, 16-12-1991, 17-12-1991, 18-12-1991, 20-12-1991, 23-12-1991, 24-12-1991, 25-12-1991, 26-12-1991, 27-12-1991, 28-12-1991, 31-12-1991, 01-01-1992, 02-01-1992, 09-01-1992, 10-01-1992, 11-01-1992, 12-01-1992, 19-01-1992, 20-01-1992, 21-01-1992, 22-01-1992, 29-01-1992, 01-02-1992, 02-02-1992]

Without leading 0s for days/months:

[20-11-1991, 23-11-1991, 24-11-1991, 25-11-1991, 26-11-1991, 27-11-1991, 28-11-1991, 30-11-1991, 3-12-1991, 4-12-1991, 5-12-1991, 6-12-1991, 7-12-1991, 8-12-1991, 10-12-1991, 13-12-1991, 14-12-1991, 15-12-1991, 16-12-1991, 17-12-1991, 18-12-1991, 20-12-1991, 23-12-1991, 24-12-1991, 25-12-1991, 26-12-1991, 27-12-1991, 28-12-1991, 31-12-1991, 3-1-1992, 4-1-1992, 5-1-1992, 6-1-1992, 7-1-1992, 8-1-1992, 10-1-1992, 13-1-1992, 14-1-1992, 15-1-1992, 16-1-1992, 17-1-1992, 18-1-1992, 20-1-1992, 23-1-1992, 24-1-1992, 25-1-1992, 26-1-1992, 27-1-1992, 28-1-1992, 31-1-1992]

Challenge rules:

  • The input and output dates may be in any reasonable (date-)format. Can be as a string in any dMy format (including optional separators), list of three integers, your language's native Date-object, etc. Output may be a list/array/stream, printed to STDOUT, a single delimited String, etc.
  • You are allowed to include or exclude leading 0s for days/months in your outputs. Please specify which of the two you use in your answer, since it will cause different results. I.e. 1-1-1991 has 2 unique digits, but 01-01-1991 as 3 unique digits.
  • You don't have to deal with leap years and differences of Gregorian vs Julian calendars. You can assume the date-ranges given in the test cases will never go over February 28th/March 1st for years divisible by 4.
  • The input-digit \$n\$ is guaranteed to be in the range \$[1,8]\$, so dealing with \$n=0\$ is unspecified (returning an empty list would be most reasonable, but giving an error or incorrect result is fine as well; you won't have to deal with that input).

General rules:

  • This is , so shortest answer in bytes wins.
    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
  • Default Loopholes are forbidden.
  • If possible, please add a link with a test for your code (i.e. TIO).
  • Also, adding an explanation for your answer is highly recommended.

Test cases:

Inputs: [12-11-1991, 02-02-1992], 4
Outputs with leading 0s:    [20-11-1991, 23-11-1991, 24-11-1991, 25-11-1991, 26-11-1991, 27-11-1991, 28-11-1991, 30-11-1991, 01-12-1991, 02-12-1991, 09-12-1991, 10-12-1991, 13-12-1991, 14-12-1991, 15-12-1991, 16-12-1991, 17-12-1991, 18-12-1991, 20-12-1991, 23-12-1991, 24-12-1991, 25-12-1991, 26-12-1991, 27-12-1991, 28-12-1991, 31-12-1991, 01-01-1992, 02-01-1992, 09-01-1992, 10-01-1992, 11-01-1992, 12-01-1992, 19-01-1992, 20-01-1992, 21-01-1992, 22-01-1992, 29-01-1992, 01-02-1992, 02-02-1992]
Outputs without leading 0s: [20-11-1991, 23-11-1991, 24-11-1991, 25-11-1991, 26-11-1991, 27-11-1991, 28-11-1991, 30-11-1991, 3-12-1991, 4-12-1991, 5-12-1991, 6-12-1991, 7-12-1991, 8-12-1991, 10-12-1991, 13-12-1991, 14-12-1991, 15-12-1991, 16-12-1991, 17-12-1991, 18-12-1991, 20-12-1991, 23-12-1991, 24-12-1991, 25-12-1991, 26-12-1991, 27-12-1991, 28-12-1991, 31-12-1991, 3-1-1992, 4-1-1992, 5-1-1992, 6-1-1992, 7-1-1992, 8-1-1992, 10-1-1992, 13-1-1992, 14-1-1992, 15-1-1992, 16-1-1992, 17-1-1992, 18-1-1992, 20-1-1992, 23-1-1992, 24-1-1992, 25-1-1992, 26-1-1992, 27-1-1992, 28-1-1992, 31-1-1992]

Inputs: [19-09-2019, 30-09-2019], 5
Outputs (same with and without leading 0s): [23-09-2019, 24-09-2019, 25-09-2019, 26-09-2019, 27-09-2019, 28-09-2019, 30-09-2019]

Inputs: [19-09-2019, 30-09-2019], 8
Output (same with and without leading 0s): []

Inputs: [20-06-1749, 30-06-1749], 8
Outputs with leading 0s:    [23-06-1749, 25-06-1749, 28-06-1749]
Outputs without leading 0s: []

Inputs: [10-12-1969, 12-01-1970], 6
Outputs (same with and without leading 0s): [30-12-1969]

Inputs: [10-12-1969, 12-01-1970], 5
Outputs with leading 0s:    [10-12-1969, 13-12-1969, 14-12-1969, 15-12-1969, 17-12-1969, 18-12-1969, 20-12-1969, 23-12-1969, 24-12-1969, 25-12-1969, 27-12-1969, 28-12-1969, 31-12-1969, 02-01-1970, 03-01-1970, 04-01-1970, 05-01-1970, 06-01-1970, 08-01-1970, 12-01-1970]
Outputs without leading 0s: [10-12-1969, 13-12-1969, 14-12-1969, 15-12-1969, 17-12-1969, 18-12-1969, 20-12-1969, 23-12-1969, 24-12-1969, 25-12-1969, 27-12-1969, 28-12-1969, 31-12-1969, 2-1-1970, 3-1-1970, 4-1-1970, 5-1-1970, 6-1-1970, 8-1-1970, 12-1-1970]

Inputs: [11-11-1111, 11-11-1111], 1
Output (same with and without leading 0s): [11-11-1111]
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1
  • \$\begingroup\$ I know it is specifically not the format demanded here, but I just comment here to "spread the word": there is an internationnal standard that programmers should try to use for dates (in logs, filenames, etc) : en.wikipedia.org/wiki/ISO_8601. (uses a 'T' to desambiguate (without the T the date+time could be something else entirely) and specifies the separators to use, and when one can not use separators [ie proposes different variant (from short to long] of the same standard]: the main one being: YYYY-MM-DDThh:mm:ss.mmm +hh:mm, the +hh:mm being your locale's time offset from UTC .) \$\endgroup\$
    – Olivier Dulac
    Commented Oct 10, 2019 at 9:58

10 Answers 10

Reset to default
5
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R, 81 bytes

function(s,e,n,x=seq(s,e,1))x[lengths(sapply(strsplit(paste(x),""),unique))==n+1]

Try it online!

Uses R’s native date format and has leading zeros on day and month.

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3
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Red, 93 bytes

func[a b n][until[if n = length? exclude d: rejoin[a/4"-"a/3"-"a/2]"-"[print d]b < a: a + 1]]

Try it online!

Without leading 0s for days/months.

Too bad that Red converts internally 09-10-2019 to 9-Oct-2019 - that's why I need to extract the day/month/year individually.

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2
  • \$\begingroup\$ I'm not familiar with Red but it seems like there's a lot of whitespace that could be cut out of this. Pardon me if I'm wrong, though. \$\endgroup\$
    – moltarze
    Commented Oct 10, 2019 at 9:51
  • \$\begingroup\$ @connectyourcharger No problem! Seemingly there are several unnecessary whitespaces, but they are needed to separate the tokens. Words (identifiers) in Red can include -=+*<>?!~&, that's why a whitespace (or (...)[...]"...") are needed. \$\endgroup\$
    – Galen Ivanov
    Commented Oct 10, 2019 at 10:35
2
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Japt, 23 bytes

Takes the date inputs as Unix timestamps, outputs an array of strings with formatting and leading 0s dependent on your locale. Would be 1 byte shorter in Japt v2 but there seems to be a bug when converting Date objects to strings.

òV864e5@ÐX s7Ãf_¬â ʶWÄ

Try it

òV864e5@ÐX s7Ãf_¬â ʶWÄ     :Implicit input of integers U=s,V=e & W=n
òV                          :Range [U,V]
  864e5                     :  With step 86,400,000 (24*60*60*1000)
       @                    :Map each X
        ÐX                  :  Convert to Date object
           s7               :  Convert to locale date string
             Ã              :End map
              f             :Filter
               _            :By passing each through the following function
                ¬           :  Split
                 â          :  Deduplicate
                   Ê        :  Length
                    ¶       :  Is equal to
                     WÄ     :  W+1, to account for the inclusion of the delimiting "/" or "-"
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2
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Python 3.8 (pre-release), 84 bytes

-6 bytes thanks to Gloweye

lambda s,e,n:[d for i in range((e-s).days+1)if-len(set(d:=str(s+type(e-s)(i))))==~n]

An unnamed function which returns a list of strings (counting/including leading zeros) that accepts three arguments:

  • s, the start - a datetime.date object;
  • e, the end - a datetime.date object; and
  • n, the number of days - an int object.

Try it online!

Note: As the function accepts datetime.date objects I have not counted the import code for that (and have worked around importing the datetime.timedelta object as it is indirectly accessible via subtraction of these input objects).

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4
  • 1
    \$\begingroup\$ type(obj) is 3 bytes shorter than obj.__class__. \$\endgroup\$
    – Gloweye
    Commented Oct 10, 2019 at 8:15
  • \$\begingroup\$ @Gloweye awesome thanks! \$\endgroup\$
    – Jonathan Allan
    Commented Oct 10, 2019 at 9:07
  • \$\begingroup\$ @Gloweye saves 6 in the end (the dot goes and the brackets are already present) \$\endgroup\$
    – Jonathan Allan
    Commented Oct 10, 2019 at 9:14
  • 1
    \$\begingroup\$ I hadn't analyzed to that depth yet, just saw the __class__ and made a quick comment. Always glad to help. \$\endgroup\$
    – Gloweye
    Commented Oct 10, 2019 at 9:15
2
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Ruby -rdate, 54 bytes

Takes 2 Date objects and a number as input, and returns a list of Date objects as output. Handles leap years and uses leading zeroes.

->a,b,n{(a..b).select{|d|d.to_s.chars.uniq.size==n+1}}

Try it online!

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1
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JavaScript (ES6), 91 bytes

Takes input as (n)(end)(start), where the dates are expected as Unix timestamps in milliseconds. Returns a space-separated list of dates in format yyyy-mm-dd.

Leading 0s are included.

n=>b=>g=a=>a>b?'':(new Set(d=new Date(a).toJSON().split`T`[0]).size+~n?'':d+' ')+g(a+864e5)

Try it online!

Improve this answer
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1
  • 1
    \$\begingroup\$ This snippet will probably fail if DST shifted happened (a+864e5 may not be tomorrow in that case). But thankfully, tio use UTC timezone which doesn't have DST. -- from someone who use the same logic in product website and realize something wrong until DST shifted... \$\endgroup\$
    – tsh
    Commented Oct 10, 2019 at 9:37
1
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PHP, 90 bytes

for([,$s,$e,$n]=$argv;$s<=$e;)$n-count(count_chars($d=date(Ymd,86400*$s++),1))||print$d._;

Try it online!

This is with leading 0s. Inputs are command arguments ($argv) and dates are Unix timestamps in days (basically standard seconds / 86400), I used this format as we don't need the time in this challenge and it allowed me to golf 1 more byte. Keeps adding a day to start until it reaches the end and prints any dates with $n unique digits in it, separated by _ in Ymd format.

Also have a 89 bytes alternative which prints dates to output in same format as input (Unix timestamps in days).

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1
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Java (JDK), 86 bytes

(s,e,n)->s.datesUntil(e.plusDays(1)).filter(d->(""+d).chars().distinct().count()==n+1)

Try it online!

I chose to use leading 0s.

Credits

  • -24 thanks to Kevin Cruijssen who didn't know he could golf away that much :p
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3
  • 1
    \$\begingroup\$ Maybe I shouldn't have made it inclusive on both ends, then you could have used Java 9's datesUntil for 103 bytes. ;) Nice answer. I don't see anything that could be golfed personally. \$\endgroup\$
    – Kevin Cruijssen
    Commented Oct 10, 2019 at 11:54
  • 1
    \$\begingroup\$ @KevinCruijssen Well, actually, you gave an extremely nice way of golfing! Just add .plusDays(1) and remove the .forEach(System.out::println) and it's a very much golfed answer because as you wrote, dates can be returned as value objects and streams are allowed. ;-) I had no clue that datesUntil even existed! Thank you for that :-) \$\endgroup\$
    – Olivier Grégoire
    Commented Oct 10, 2019 at 12:34
  • \$\begingroup\$ Woops, forgot about my own rules allowing stream returns, haha XD I'm an idiot. But glad I could help you with my idiocy. ;p \$\endgroup\$
    – Kevin Cruijssen
    Commented Oct 10, 2019 at 13:46
1
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C# (Visual C# Interactive Compiler)

Without leading 0s, 104, 103 bytes

(s,e,n)=>new int[(e-s).Days+1].Select((x,i)=>s.AddDays(i)).Where(x=>$"{x:yyyMd}".Distinct().Count()>=n)

Try it online!

With leading 0s, 106 105 bytes

(s,e,n)=>new int[(e-s).Days+1].Select((x,i)=>s.AddDays(i)).Where(x=>$"{x:yyyMMdd}".Distinct().Count()>=n)

Try it online!

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2
  • \$\begingroup\$ You can remove the space at x =>$" in both your versions for -1. :) \$\endgroup\$
    – Kevin Cruijssen
    Commented Oct 15, 2019 at 12:29
  • \$\begingroup\$ @KevinCruijssen ah fuck thanks x) I'm sure someone can come up with a shorter solution too, I'm not satisfied with this one \$\endgroup\$
    – Innat3
    Commented Oct 15, 2019 at 12:35
0
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Kotlin, 119 bytes

fun f(a:LocalDate,b:LocalDate,c:Long)=a.datesUntil(b.plusDays(1)).filter{it.toString().chars().distinct().count()==c+1}

Without leading 0s, takes in two java.time.LocalDate and a Long, returns a Stream of LocalDates

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1
  • \$\begingroup\$ Hi, welcome to CCGC! Could you perhaps add a Try it online link with test code to verify it works? Also, I don't know Kotlin, but is it possible to replace it.toString() with (it+"") to save a few bytes? I know this is possible in some other languages like Java or .NET C#. \$\endgroup\$
    – Kevin Cruijssen
    Commented Oct 15, 2019 at 12:31

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