Daily Challenge #291 - Extended Weekend

If the first day of the month is a Friday, it is likely that the month will have an Extended Weekend. That is to say, it could have five Fridays, five Saturdays and five Sundays.

In this Kata, you will be given a start year a and an end year b(1000 <= a <= b <= 275760). Your task will be to find months that have extended weekends and return:

The first and last month in the range that has an extended weekend
The number of months that have extended weekends in the range, inclusive of start year and end year.
If, there is no Extended Weekend, just return ["","",0].

Example

solve(2016,2020) = ("Jan","May",5).
The months are: Jan 2016, Jul 2016, Dec 2017, Mar 2019, May 2020

Tests

solve(2016,2020)
solve(1900,1950)
solve(1800,2500)
solve(1000,275760)

Good luck!

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This challenge comes from myjinxin2015 on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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Comments

['']

• To create a range as an array in JS with inclusive borders , you can use Array(end + 1).keys , then slice it from the start number.
• For extended month, only look for months which takes 31 days.
• In JS Date object, month index starts from 0. Jan => 0 , Feb => 1 and so on.
• In JS Date object, week day index starts from 0 with Sunday, so Friday's index is 5.

const solve = (a,b) => {
 const FRIDAY = 5, JAN = 0, MAR = 2 , MAY = 4, JUL = 6, AUG = 7, OCT = 9, DEC = 11
 const months = ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]
 const extendedMonths = [...Array(b+1).keys()].slice(a).map(year => ([JAN,MAR,MAY,JUL,AUG,OCT,DEC].filter(monthIndex => new Date(year,monthIndex,1).getDay() === FRIDAY)).map(monthIndex => `${months[monthIndex]} ${year}`)).flat()
 const firstExtendedMonth = extendedMonths.length > 0 ? extendedMonths[0].split(" ")[0] : ""
 const lastExtendedMonth = extendedMonths.length > 0 ? extendedMonths[extendedMonths.length - 1].split(" ")[0] : ""
 return [firstExtendedMonth,lastExtendedMonth,extendedMonths.length]
}

[Yasuhiro Yamada]

Solution in Bash and some general commands.

#!/bin/bash
solve () {
  seq $1 $2 |
    sed -n -e'h;s/$/'{01,03,05,07,08,10,12}'01/;p;x' |
    LANG=C date -f- +'%b %a' |
    grep Fri |
    awk 'NR==1{printf $1}END{print " "$1" "NR}'
}

solve 2016 2020   # => Jan May 5
solve 1900 1950   # => Mar Dec 51
solve 1800 2500   # => Aug Oct 702
solve 1000 275760 # => Aug Aug 274761
solve 1973 1973 #=> [empty] [empty] 0