Daily Challenge #27 - Unlucky Days

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Daily Challenge (132 Part Series)

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(or not...) 77) Daily Challenge #77 - Bird Mountain 78) Daily Challenge #78 - Number of Proper Fractions with Denominator d 79) Daily Challenge #79 - Connect Four 80) Daily Challenge #80 - Longest Vowel Change 81) Daily Challenge #81 - Even or Odd 82) Daily Challenge #82 - English Beggars 83) Daily Challenge #83 - Deodorant Evaporator 84) Daily Challenge #84 - Third Angle of a Triangle 85) Daily Challenge #85 - Unwanted Dollars 86) Daily Challenge #86 - Wouldn't, not Would. 87) Daily Challenge #87 - Pony Express 88) Daily Challenge #88 - Recursive Ninjas 89) Daily Challenge #89 - Extract domain name from URL 90) Daily Challenge #90 - One Step at a Time 91) Daily Challenge #91 - Bananas 92) Daily Challenge #92 - Boggle Board 93) Daily Challenge #93 - Range Extraction 94) Daily Challenge #94 - Last Digit 95) Daily Challenge #95 - CamelCase Method 96) Daily Challenge #96 - Easter Egg Crush Test 97) Daily Challenge #97 - Greed is Good 98) Daily Challenge #98 - Make a Spiral 99) Daily Challenge #99 - Balance the Scales 100) Daily Challenge #100 - Round Up 101) Daily Challenge #101 - Parentheses Generator 102) Daily Challenge #102 - Pentabonacci 103) Daily Challenge #103 - Simple Symbols 104) Daily Challenge #104 - Matrixify 105) Daily Challenge #105 - High-Sum Matrix Drop 106) Daily Challenge #106 - Average Fuel Consumption 107) Daily Challenge #107 - Escape the Mines 108) Daily Challenge #108 - Find the Counterfeit Coin 109) Daily Challenge #109 - Decorate with Wallpaper 110) Daily Challenge #110 - Love VS. 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Friday 13th or Black Friday is considered as an unlucky day. Calculate how many unlucky days are in the given year.

Can you find the number of Friday 13th in the given year? Good luck!

Input: Year as an integer.

Output: Number of Black Fridays in the year as an integer.

Examples:

unluckyDays(2015) == 3
unluckyDays(1986) == 1

Note: In Ruby years will start from 1593.


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

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markdown guide
 

JavaScript

const unluckyDays = year => {
  let unlucky = 0;
  for (x = 0; x < 12; x++) {
    unlucky += new Date(year, x, 13).getDay() === 5 ? 1 : 0;
  }
  return unlucky;
}

Live demo on CodePen

 

Brilliant solution, very simple and effective. Although I think ? 1 : 0 is not necessary :p

 

Yes. It's not really necessary because true is turn into 1, and false to 0. I have a second version using that and a reducer in the demo.

 

Javascript shorty

matchingDayCount = (year, dayOfMonth=13, dayOfWeek=5) => 
  [...Array(12)].reduce((acc, _, month) =>
    acc + Number(new Date(year, month, dayOfMonth).getDay() == dayOfWeek), 0)

[2015, 1986].map(y => matchingDayCount(y)).join() // >> 3,1
 

Javascript:

function unluckyDays(year) {
    return [[2, 1, 3, 1, 1, 2, 2], [1, 2, 2, 1, 1, 3, 2]]
    [+(year % 400 === 0 || year % 4 === 0 && year % 100 > 0)]
    [new Date(year, 0, 13).getDay()]
}

You can determine the number of Fridays in a given year as the number of days between the 13th of one and the next month is fixed. All you need to know is the day of January 13th and whether the year is a leap-year.

 

In one single Smalltalk line, because we can:

2015 asYear months count: [ :m | (m start asDate + 12 days) dayOfWeek = 6 ]. "return 3"
1986 asYear months count: [ :m | (m start asDate + 12 days) dayOfWeek = 6 ]. "return 1"
 

Perl solution, using the core library Time::Piece.

#!/usr/bin/perl
use warnings;
use strict;

use Time::Piece;

sub unlucky_days {
    my ($year) = @_;
    return grep $_->fullday eq 'Friday',
           map 'Time::Piece'->strptime("$year-$_-13", '%Y-%m-%d'),
           1 .. 12
}

use Test::More tests => 2;
is unlucky_days(2015), 3, 'year 2015';
is unlucky_days(1986), 1, 'year 1986';

It works because grep in scalar context returns the number of trues.

 

Elixir:

defmodule Unlucky do
  def days(year) do
    1..12
    |> Enum.map(&Date.from_erl!({year, &1, 13}))
    |> Enum.filter(&(Date.day_of_week(&1) == 5))
    |> Enum.count()
  end
end
 
require "date"

def unlucky_days(year)
  unlucky = 0
  1.upto(12).each do |month|
    date = Date.new(year, month, 13)
    unlucky += 1 if date.wday == 5
  end
  unlucky
end
 

My python sol :

def unlucky_days(year):
    count = 0
    cal = Calendar()
    for month in range(1,13):
        for day, week_day in cal.itermonthdays2(year, month):
            if day == 13 and week_day == 4:
                count += 1
    return count
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