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Introduction

In the United States, national elections are normally held on the first Tuesday after the first Monday in November.

Challenge

Write a program that takes a Year as integer, and a pair of (day of week as character and ordinal as integer) and output the date as "yyyy-mm-dd" Remember, the second pair is after the first pair. use this for inspiration: https://codegolf.stackexchange.com/questions/178227/get-the-date-of-the-nth-day-of-week-in-a-given-year-and-month[][1]

Example Input and Output

Example input:

  • 2021 November Tuesday 1 Monday 1
  • 2020 November Tuesday 1 Monday 1
  • 2020 December Friday 2 Wednesday 1
  • 2019 April Wednesday 1 Friday 1
  • 2023 February Tuesday 4 Wednesday 1
  • 2023 February Wednesday 5 Thursday 1
  • 2022 December Sunday 5 Sunday 55

Expected output:

  • "2021-11-02"
  • "2020-11-03"
  • "2020-12-11"
  • "2019-04-10"
  • "2023-02-28"
  • "NA"
  • "NA"

Note: the output indicated as "NA" could be any negative finding such as null, but not an error. [1]: Get the date of the nth day of week in a given year and month

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  • 11
    \$\begingroup\$ Hi and welcome! This is an interesting challenge, but please avoid cumbersome I/O formats, and for next time, I highly recommend using the sandbox. \$\endgroup\$
    – Adám
    Feb 1 at 22:22
  • 1
    \$\begingroup\$ Suggested test case: 2023 February Tuesday 4 Wednesday 1 \$\endgroup\$
    – Adám
    Feb 1 at 22:27
  • 1
    \$\begingroup\$ Suggested test case: 2023 February Wednesday 5 Thursday 1 \$\endgroup\$
    – Adám
    Feb 1 at 22:28
  • 1
    \$\begingroup\$ Suggested test case: 2022 December Sunday 5 Sunday 55 \$\endgroup\$
    – Adám
    Feb 1 at 22:29
  • 2
    \$\begingroup\$ @gildux - The example that you ask about, 2021 November Monday 1 Tuesday 1 is asking for "the first Tuesday that is after the first Monday in November, 2021". You should mentally parse it as (2021 November) (Monday 1) (Tuesday 1) \$\endgroup\$ Feb 2 at 11:25

4 Answers 4

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JavaScript, 102 93 91 88 80 78 77 bytes

Oof, this made my brain itch. Possibly still some room for improvement.

(m,t,x=0)=>g=w=>(d=new Date(...m,++x)).getDay()-w[0]||--w[1]?g(w):w!=t?g(t):d

Try it online!

Call with f([y,m],[W,O])([w,o]), where:

  • y is the year,
  • m is the 0-indexed month,
  • W is the 0-indexed target weekday (0=Sunday),
  • O is the ordinal of W,
  • w is the 0-indexed starting weekday, and,
  • o is the ordinal of w

EDIT: Looking at the rest of the solutions coming in, they seem to be interpreting the I/O requirements of the spec a lot more strictly than I am. After a few read-throughs, I'm not seeing that strictness, nor would it be reasonable to assume it's that strict. Still, though, seeing as I appear to be the only one taking the requirements this loosely, I'll take a stab at an alternative solution on Tuesday. (It's a bank holiday weekend in Ireland)

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  • \$\begingroup\$ Based on the last two test cases, if the calculated date exceeds the input year+month, it should output NA. But maybe it's good to verify with OP whether this is optional or mandatory, since just calculating beyond the current year+month works perfectly fine for larger input. 🤷 \$\endgroup\$ Feb 2 at 17:02
  • 1
    \$\begingroup\$ Hopefully, "undefined behaviour" is sufficient in those cases, @KevinCruijssen. \$\endgroup\$
    – Shaggy
    Feb 2 at 17:11
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C (gcc), 439 428 bytes

  • -11 thanks to ceilingcat

Most of the code is parsing the input.

#define z(d)(*d-83?*d-84?*d-77?*d-87?5:3:1:d[1]-117?4:2:d[1]-117?6:0)
#define Z(d,e)(d-e+(d<e?7:0))
t[20],a,b,n,x,y;char*m,*d,*e;f(s){sscanf(s,"%d %ms %ms %d %ms %d",&y,&m,&d,&a,&e,&b);t[5]=y-=1900;x=*m-78;t[4]=n=x+4?~x?x+13?x+8?x-5?x-1?x?11:10:9:8:1:m[1]-112?7:3:m[2]-114?2:4:m[1]-97?m[2]-110?6:5:0;t[3]=1;mktime(&t);t[3]+=Z(z(e),t[6])+(a+b-2)*7+Z(z(d),z(e));mktime(&t);t[4]-n|t[5]-y||printf("%04d-%02d-%02d",y+1900,n+1,t[3]);}

Try it online!

Ungolfed:

// Day of week: Su=0, M=1, Tu=2, W=3, Th=4, F=5, Sa=6
#define z(d)(*d-'S'?*d-'T'?*d-'M'?*d-'W'?5:3:1:d[1]-'u'?4:2:d[1]-'u'?6:0)
// Add one week if the target day of week precedes the current one
#define Z(d,e)(d-e+(d-e<0?7:0))
t[20], // struct tm for time conversions (important parts are initialized each run through)
a, b, // week numbers
x, // temporary for month calculation
n, y; // month and year
char *m, *d, *e; // original string versions of days, month and year
f(int*s) {
  sscanf(s,"%d %ms %ms %d %ms %d",&y,&m,&d,&a,&e,&b); // get the parsed data
  t[5]=y-=1900; // adjust the year
  x=*m-78; // get most values into single-digit
  // Month: Ja=0 F=1 M_r=2 Ap=3 M_y=4 J_n=5 J_l=6 Au=7 S=8 O=9 N=10 (else) 11
  //t[4]=n=*m-'J'?*m-'M'?*m-'A'?*m-'F'?*m-'S'?*m-'O'?*m-'N'?11:10:9:8:1:m[1]-'p'?7:3:m[2]-'r'?2:4:m[1]-'a'?m[2]-'n'?6:5:0;
  t[4]=n=x+4?~x?x+13?x+8?x-5?x-1?x?11:10:9:8:1:m[1]-112?7:3:m[2]-114?2:4:m[1]-97?m[2]-110?6:5:0;
  t[3]=1; // start on the first day of the month
  mktime(&t); // get the day of week information
  t[3]+=Z(z(e),t[6])+(a+b-2)*7+Z(z(d),z(e)); // add offsets
  mktime(&t); // convert to canonical form
  t[4]-n|t[5]-y?printf("%04d-%02d-%02d",y+1900,n+1,t[3]); // only print date if it is still inside the month
}
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05AB1E, 204 (or 126?) bytes

With strict I/O and outputting NA when the date goes beyond the given input month+year: 204 bytes:

"Y`т‰0Kθ4ÖUD<i\28X+ë<7%É31α}‹iY¬>0ëY1¾ǝDÅsD12‹i>1ë\1Dǝ¤>2}}ǝV"ˆ$”‚應…ä†ï€¿…Ë…ê†Ä…æ…Ì…Í……”#s#©1èk>Dˆ®н)V®2ô¦RvyθFNĀi¯н.V}[Y`UD3‹©12*+>₂*T÷®Xα©т%D4÷®т÷©4÷®·(O7%”ŒÍ‹Ó‹ŽŒ¹ŒêŒÛŠ¯”#yнkQ#¯н.V]YÂT‰J'-ýsÅs¯θÊi„NA

The cumbersome I/O and 05AB1E's lack of date builtins make for a pretty big program. 🙃

Try it online or verify all test cases.

With flexible I/O and just continue calculating instead of outputting NA: 126 bytes:

"Y`т‰0Kθ4ÖUD<i\28X+ë<7%É31α}‹iY¬>0ëY1¾ǝDÅsD12‹i>1ë\1Dǝ¤>2}}ǝV"ˆ1šVvyθFNĀi¯н.V}[Y`UD3‹©12*+>₂*T÷®Xα©т%D4÷®т÷©4÷®·(O7%yнQ#¯н.V]Y

Two inputs in the formats [month,year] and [[weekday2,amount2],[weekday1,amount1]], where month is an integer 1-12 and weekday are integers 0-6 for Saturday to Friday.
Outputs in the format [day,month,year].

Still pretty big with the manual calculations, but apparently more than 1/3rd of my top program's bytes are to deal with just the I/O formats.

Try it online.

Explanation (of the larger strict I/O version):

Step 0: Create a function to go to the next day, which we'll re-use later on (see this 05AB1E answer of mine for an in-depth explanation of how we're going to the next day manually of a given date):

"Y`т‰0Kθ4ÖUD<i\28X+ë<7%É31α}‹iY¬>0ëY1¾ǝDÅsD12‹i>1ë\1Dǝ¤>2}}ǝV"
                 # Push this string to act as function later on with an eval
 ˆ               # Add it to the global array†

† The reason I use the global array instead of a variable: I'm already using all three U/X, V/Y, and ©® variable setters/getters in the actual manual date calculations.

Step 1: Extract the year and month from the input, and parse it as [1,m,y] triplet-list:

$                # Push 1 and the input-string
 ”‚應…ä†ï€¿…Ë…ê†Ä…æ…Ì…Í……”
                 # Push dictionary string "January February March April May June July August September October November December"
  #              # Split it on spaces to a list
   s             # Swap so the input-string is at the top of the stack
    #            # Split it on spaces as well
     ©           # Store this sextuple input-list in variable `®` (without popping)
      1è         # Pop and get the Month at (0-based) index 1
        k        # Get the (0-based) index of this month in the list
         >       # Increase it by 1 to a 1-based index
          Dˆ     # Add a copy to the global array as well
            ®    # Push the input-list of variable `®` again
             н   # Pop and leave its first item (the year)
              )  # Wrap all three values into a list: [1,m,y]
               V # Pop and store it in variable `Y`

Step 2: Parse the remainder of the input-string, and start looping:

®                # Push the input-list from variable `®` again
 2ô              # Split it into parts of size 2
   ¦             # Remove the first part (the [year,"Month"])
    R            # Reverse the other two parts
     v           # For each over the pairs `y` in the pair:
      yθ         #  Push the last item of the current pair (the amount of days)
        F        #  Pop and inner loop that many times:
         NĀi     #   If it's NOT the first iteration:
            ¯н.V #    Go to the next day
                 #    by evaluating the first item of the global array of step 0
           }     #   Close the if-statement
         [       #   Start an inner infinite loop:

Step 3a: Calculate the DayOfWeek of the current date as [0,1,2,3,4,5,6] for [Saturday,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday] respectively (see again this 05AB1E answer of mine for an in-depth explanation of how I calculate the Day of the Week manually of a given date):

Y`UD3‹©12*+>₂*T÷®Xα©т%D4÷®т÷©4÷®·(O7%

Step 3b+c: If this day is equal to the input-date of the current pair y, stop the infinite loop. If not, go to the next day and continue looping.

”ŒÍ‹Ó‹ŽŒ¹ŒêŒÛŠ¯” #    Push dictionary string "Saturday Sunday Monday Tuesday Wednesday Thursday Friday"
 #               #    Split it on spaces to a list
  yн             #    Push the first item of the current pair (the day)
    k            #    Get its 0-based index in the list of weekdays
     Q           #    Check whether it's equal to the calculated DayOfWeek from step 3a
      #          #    If they're equal:
      #          #     Stop the inner infinite loop
       ¯н.V      #    (Else) Go to the next day by evaluating step 0 again

Step 4: Format the resulting date to the desired output-format:

]                # Close the three loops
 Y               # Push the resulting date `Y`
  Â              # Bifurcate this triplet; short for Duplicate & Reverse copy
   T‰J           # Format the day/month with leading 0:
   T‰            #  Divmod each inner value by 10
     J           #  Join each inner pair together
      '-ý       '# Join the triplet-list with "-" delimiter
  s              # Swap the triplet to the top of the stack again
   Ås            # Only leave its middle item (the month)
     ¯θ          # Push the last item of the global array (the parsed input-month)
       Êi        # If they are NOT equal:
         „NA     #  Push string "NA"
                 # (after which the top of the stack is output implicitly as result)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why ”‚應…ä†ï€¿…Ë…ê†Ä…æ…Ì…Í……” is "January February March April May June July August September October November December" and ”ŒÍ‹Ó‹ŽŒ¹ŒêŒÛŠ¯” is "Saturday Sunday Monday Tuesday Wednesday Thursday Friday".

Both "ˆ$”‚應…ä†ï€¿…Ë…ê†Ä…æ…Ì…Í……”#I#1èk> and ”ŒÍ‹Ó‹ŽŒ¹ŒêŒÛŠ¯”#yнk can probably be golfed a bit with magic numbers of some sort, but since I suck at those and I also couldn't really be bothered in an answer already this big because of manual date calculations, the indexing into dictionary strings will do for now.

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  • \$\begingroup\$ OP has confirmed that any undefined behaviour, except an error, is acceptable for the "NA" test cases. \$\endgroup\$
    – Shaggy
    Feb 2 at 17:50
  • \$\begingroup\$ @Shaggy His comment to me contradicts that lol.. "it should not return a date and should not result in an error". But I'll just look at his reply to you. ;) \$\endgroup\$ Feb 2 at 17:59
  • \$\begingroup\$ @KevinCruijssen these comments do not contradict. consider them together. \$\endgroup\$ Feb 2 at 18:51
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Go, 262 bytes

import."time"
func f(y,m,d,e string,n,o int)string{t,_:=Parse("2006January",y+m)
D,i,j:=Hour*24,1,1
for t.Weekday().String()!=e{t=t.Add(D)}
for;j<o;j++{t=t.Add(D*7)}
for t.Weekday().String()!=d{t=t.Add(D)}
for;i<n;i++{t=t.Add(D*7)}
return t.Format("2006-01-02")}

Attempt This Online!

Extremely similar to my solution for "Get the date of the nth day of week in a given year and month".

For the NA test cases, it continues on to the next month/year until it finds the n-th dday after the o-th eday.

Explanation

import."time"
func f(y,m,d,e string,n,o int)string{
t,_:=Parse("2006January",y+m)
D:=Hour*24

Some initial setup: getting the 1st of month m in the year y, and defining a Day variable.

for t.Weekday().String()!=e{t=t.Add(D)}
for j:=1;j<o;j++{t=t.Add(D*7)}

Skip forward to the o-th eday.

for t.Weekday().String()!=d{t=t.Add(D)}
for i:=1;i<n;i++{t=t.Add(D*7)}

Skip forward to the n-th dday.

return t.Format("2006-01-02")}

Return a formatted date.

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