Tesco's Burger Relish Best Before End date number

Question

Given a date between 2010-01-01 and 2099-12-31 as three integers (please state your order if not [year,month,day]), answer with Tesco's corresponding five-digit Burger Relish Best Before End date number.

The format is simply the last two digits of the year, followed immediately by three digits representing the day of that year, between 000 and 365.

Cookies if you can do this using only numeric operations and comparisons. You may put such code separately from your golfed code, if it isn't also the shortest code you have.

Test cases

[year,month,day] → BRBBED:

[2010,1,1] → 10000

[2019,7,5] → 19185

[2020,5,20] → 20140

[2096,12,31] → 96365

[2099,12,31] → 99364

Answer 1

PowerShell, 74 70 67 bytes

param($y,$m,$d)-join"$y 00$((date "$m/$d/$y"|% d*r)-1)"[2,3+-3..-1]

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Exactly what it says on the tin (but formatted weirdly). Takes in $year, $month, $day, plucks out the last two digits of the $year, gets a .NET datetime object of the specified day, then gets the dayofyear (with |% d*r) thereof. Subtracts one to make it zero-indexed, then uses string formatting and slicing to make it three-padded (e.g., 000 instead of 0), and then -joins it all together into a single string with implicit output.

Note this is culture dependent due to the datetime formatting. This works in en-us, which is what TIO also uses.

-3 bytes thanks to mazzy

Answer 2

C# (Visual C# Interactive Compiler), 88 78 73 71 59 57 bytes

a=>a[0]%100*1000+new DateTime(a[0],a[1],a[2]).DayOfYear-1

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Answer 3

Ruby, 40 bytes

->d{Time.gm(*d).strftime('%y%j').to_i-1}

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So close to being a built-in format, but %j is 1-indexed.

In a happier coincidence, though, Tesco Time (.gm) is one character shorter than local time (.new).

Answer 4

PowerShell, 48 45 39 bytes

Port of Expired Data's answer for C#

$args[0]*1E3+("$args"|date|% d*r)-2E6-1

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---

PowerShell, 60 58 52 bytes

-2 bytes thanks @AdmBorkBork

-join"$args 00$(("$args"|date|% d*r)-1)"[2,3+-3..-1]

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Explanation:

• The script takes three integers [year,month,day]
• "$args" makes a string like YYYY MM DD
• "YYYY MM DD"|date converts string to [DateTime]. I don't know why it works. Can you explain?
• ("YYYY MM DD"|date|% d*r)-1 calculates dayOfYear
• the script makes a result string like YYYY MM DD 00### where ### is dayOfYear
• finally the script extracts chars from positions [2,3+-3..-1] of the result string and joins this chars to result

Answer 5

Python 2, 63 61 57 bytes

lambda y,m,d:y%100*1000+(153*m-2)/5+[m+1,1>y%4][2<m]+d-33

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Returns the result as an integer.

Neil found that his Charcoal approach turned out to be lucky after some rearrangement.

Pure arithmetic and comparison solution (61 bytes):

lambda y,m,d:y%100*1000+30*m+m/2+(8<m<12)+~(0<y%4)*(2<m)+d-31

Answer 6

Python 3, 87 85 74 bytes

lambda y,m,d:sum([y*1000-2000029+28*m,3,y%4==0,3,2,3,2,3,3,2,3,2,3][:m])+d

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This is using only numeric operations and comparisons.

Answer 7

Jelly, 30 bytes

4ḍȯ2⁽¢wB+30¤_2¦ŻḣS+⁵’+³×ȷ¤_2ȷ6

A full program taking arguments y m d.
Jelly has no built-in date functionality.

Try it online!

How?

4ḍȯ2⁽¢wB+30¤_2¦ŻḣS+⁵’+³×ȷ¤_2ȷ6 - Main Link
           ¤                   - nilad followed by link(s) as a nilad:
    ⁽¢w                        -   1370
       B                       -   to binary = [1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0]
        +30                    -   add 30   = [31,30,31,30,31,30,31,31,30,31,30]
              ¦                - sparse application...
             2                 -   to indices: [2]
            _                  -   action: subtract...
4ḍ                             -     four divides y? -> 1 or 0
  ȯ2                           -     OR 2               1 or 2
               Ż               - prepend 0 = [0,31,29|28,31,30,31,30,31,31,30,31,30]
                ḣ              - head to index m
                 S             - sum
                  +⁵           - add d
                    ’          - subtract one
                         ¤     - nilad followed by link(s) as a nilad:
                      ³        -   y
                        ȷ      -   1000
                       ×       -   multiply
                     +         - add
                           2ȷ6 - 2000000
                          _    - subtract

Answer 8

Charcoal, 33 bytes

ＩΣ⟦×φ﹪θ¹⁰⁰⊖ζ÷⊕×¹⁵³⊖η⁵⎇›η²⊖⊖¬﹪θ⁴⊖η

Try it online! Link is to verbose version of code. Try it online! Link includes test suite. I don't think Charcoal has any date functions, so here's a mathematical solution. Explanation:

   ×φ﹪θ¹⁰⁰                          Year * 1000
          ⊖ζ                        Day - 1
            ÷⊕×¹⁵³⊖η⁵               Days in previous months
                     ⎇›η²            For months after Februrary
                         ⊖⊖¬﹪θ⁴     Adjust for leap years else
                               ⊖η   Adjust for short Februrary
 Σ⟦                                 Take the sum
Ｉ                                   Cast to string for implicit print

Answer 9

[R], 66 bytes

 function(a,b,c)as.numeric(strftime(paste(a,b,c,sep='-'),'%y%j'))-1

taking 3 integers and pasting them into a string strftime can recognize and extract the wanted parts. 0 indexing costs 14 bytes and another conversion to numeric

Answer 10

Ruby, 79 81 79 78 bytes

->y,m,d{(y%100*k=1000)+[d-1,l=31,y%4<1?29:28,l,s=30,l,s,l,l,s,l,s][0,m].sum%k}

Cookies version. Saved a few bytes by including the -1 and day in the array that gets summed, since those spaces were spare anyway.

EDIT: I realised that using "...%03d"%... wasn't really in the spirit of 'numerical and comparison operators only' so I've updated this answer (and gone up by 2 bytes).

EDIT-2: ... and recovered them by using the alternative sub-array index format [start,length], which in turn allowed me to combine the first two entries in the array.

EDIT-3: y%4==0 can be shortened to y%4<1

Answer 11

JavaScript, 70 57 bytes

(y,m,d,U=Date.UTC)=>y%100*1e3+(U(y,m-1,d)-U(y,0,1))/864e5

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Answer 12

Japt, 23 bytes

Takes input as 3 individual strings, assuming that's allowed.

¤+@ÐUTX ŶÐN Å}as ùT3 É

Try it

Very unhappy with this (there must be a better way to get the day of the year!) but it's been a long day. A port of darrylyeo's JS solution would be 2 bytes shorter but, for some reason, despite Japt being JavaScript, floating point inaccuracies abound.

¤+@ÐUTX ŶÐN Å}as ùT3 É     :Implicit input of strings U=y, V=d & W=d (N=[U,V,W])
¤                           :Slice the first 2 characters off U
 +                          :Append
  @                         :  Function taking an integer X as its argument
   ÐUTX                     :    Construct a Date object with U as the year, 0 as the month & X as the day
        Å                   :    Convert to date string ("yyyy-mm-dd")
         ¶                  :    Test for equality with
          ÐN Å              :    Construct a Date object from N and convert to a date string
              }             :  End function
               a            :  Return the first integer that returns true when passed through that function
                s           :  Convert to string
                  ùT3       :  Left pad with 0 to length 3
                      É     :Subtract 1

Answer 13

Red, 54 bytes

func[y m d][y % 100 * 1000 - 1 + pick to now[d m y]11]

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Answer 14

8th, 77 71 bytes

Code

needs date/utils : f third -rot d:ymd> doy n:1- swap 100 mod 1000 * + ;

Return the result on TOS

Usage

ok> 2010 1 1 f .
10000
ok> 2019 7 5 f .
19185
ok> 2020 5 20 f .
20140
ok> 2096 12 31 f .
96365
ok> 2099 12 31 f .
99364

Answer 15

Perl 6, 45 bytes

{@_.[0]%100*1000-1+Date.new(|@_).day-of-year}

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Answer 16

05AB1E, 29 26 23 bytes

¦¦₄*•ΘÏF•ºS₂+I<£O¹4ÖI<O

Try it online or verify all test cases.

Takes three loose inputs in the order \$year\$, \$month\$, \$day\$.

Explanation:

¦¦             # Remove the first two digits of the (implicit) input-year
               #  i.e. year=2024 → 24
  ₄*           # Multiply it by 1000
               #  → 24000
•ΘÏF•          # Push compressed integer 5254545
     º         # Mirror it vertically to 52545455454525
      S        # Convert it to a list of digits: [5,2,5,4,5,4,5,5,4,5,4,5,2,5]
       ₂+      # Add 26 to each: [31,28,31,30,31,30,31,31,30,31,30,31,28,31]
         I<    # Push the second input-month decremented by 1
           £   # Leave the first month-1 amount of values from the list
            O  # Sum this list
               #  i.e. month=4 → [31,28,31] → 90
¹              # Push the first input-year again
 4Ö            # Check if it's divisible by 4
               #  2024 → 1 (truthy)
I<             # Push the third input-days, decremented by 1
               #  i.e. days=3 → 2
O              # Sum everything on the stack together
               #  24000,90,1,2 → 24093
               # (which is output implicitly as result)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •ΘÏF• is 5254545.

Answer 17

Elm 0.19, 112 bytes

t y m d=1000*(y-2000)+List.sum(List.take m[d-1,31,if modBy 4 y<1 then 29 else 28,31,30,31,30,31,31,30,31,30,31])

Verify all test cases here.