US Tax Due Dates

Question

Introduction

In a standard* (e.g. not 2020) tax year, estimated payments are due quarterly according to the schedule outlined in Publication 505.

Payment Period	Due Date
January 1 – March 31	April 15
April 1 – May 31	June 15
June 1 – August 31	September 15
September 1 – December 31	January 15

* Saturday, Sunday, holiday rule, fiscal year taxpayers, and January payment exceptions can be ignored for the purpose of this question; use the standard table provided.

Challenge

Input:

A date/timestamp, either provided as input to your solution (function parameter or stdin) as a string (at a minimum specifying a specific year, month, an day of month) or native Date object (or equivalent) or the current (dynamically updated) date/time based on the clock of the executing machine.

Output:

The most specific payment period, one of Q1, Q2, Q3, or Q4 as a string, either printed to stdout/console or programmatically returned from the function, to which a timely payment made on the input date would be applied.

Input	Output
January 16 - April 15	Q1
April 16 - June 15	Q2
June 16 - September 15	Q3
September 16 - January 15	Q4

A day shall commence at 00:00 (midnight) and cover time periods up to but not including 00:00 (midnight) of the following day. The last moment of a given day, to a ms granularity is 23:59:59.999.

Answer 1

JavaScript (ES6), 53 bytes

Expects a string in "DD-MM-YYYY" format.

s=>"Q"+"4243131"[[d,m]=s.split`-`,(m*24|d>15)*9%23%7]

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39 bytes

If we can just take the month and the day.

m=>d=>"Q"+"4243131"[(m*24|d>15)*9%23%7]

Try it online!

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35 bytes

As suggested by @ovs, using a larger lookup table with a much simpler formula is actually shorter.

m=>d=>"Q"+"_4111223334444"[m+=d>15]

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Answer 2

Retina 0.8.2, 72 63 bytes

\d+
$*
1+-(1{1,5}(1)*)-1{1,15}(1)*
11$3$2$1
(111)*1+
Q$#1
0|5
4

Try it online! Link includes test cases. Takes input in YYYY-MM-DD format. Explanation:

\d+
$*

Convert to unary.

1+-(1{1,5}(1)*)-1{1,15}(1)*
11$3$2$1

Add 2, 3 or 4 to the month depending on whether the month is greater than 5 or the day is greater than 15 or not.

(111)*1+
Q$#1

Subtract 1 from the month, then integer divide it by 3 and prefix Q to the result. (This is to avoid zero width matches.)

0|5
4

Change Q0 and Q5 to Q4.

Answer 3

R, 65 62 bytes

function(d)paste0("Q",1+sum(c(3,5,8)<el(format(d-15,"%m"):1)))

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Expects input in R's native Date class. Subtract 15 days from the input, extract the month (as a string such as "04"), and convert to numeric with el(...:1) (see this tip). We then get the result by adding 1 for each of the values 3, 5, and 8 which are smaller than the result.

Answer 4

Charcoal, 28 bytes

≔Ｉ⪪Ｓ-θ§12344÷⁻§θ¹⁺›¹⁶⊟θ›⁶⊟θ³

Try it online! Link is to verbose version of code. Takes input in the form YYYY-MM-DD. Explanation:

≔Ｉ⪪Ｓ-θ

Input the date, split it on - and cast to integer.

§12344÷⁻§θ¹⁺›¹⁶⊟θ›⁶⊟θ³

Take the month, subtract 1 each if the day is less than 16 or the month is less than 6, divide by 3, then cyclically index into the string 12344, thus casting -1 and 4 to 4.

Answer 5

Python 3.8 (pre-release), 109 bytes

lambda n,i=int:'Q'+['?111223334444'[i(k:=(S:=n.split('-'))[1])],'1234'[i(k)//3-1]][k in'1469'and(i(S[0])<16)]

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Way too long.

The OP has confirmed very kindly that this is the only solution of mine which is valid.