How to resolve the algorithm Five weekends step by step in the CoffeeScript programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Five weekends step by step in the CoffeeScript programming language
Table of Contents
Problem Statement
The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.
Algorithm suggestions
Extra credit Count and/or show all of the years which do not have at least one five-weekend month (there should be 29).
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Five weekends step by step in the CoffeeScript programming language
Source code in the coffeescript programming language
startsOnFriday = (month, year) ->
# 0 is Sunday, 1 is Monday, ... 5 is Friday, 6 is Saturday
new Date(year, month, 1).getDay() == 5
has31Days = (month, year) ->
new Date(year, month, 31).getDate() == 31
checkMonths = (year) ->
month = undefined
count = 0
month = 0
while month < 12
if startsOnFriday(month, year) and has31Days(month, year)
count += 1
console.log year + ' ' + month + ''
month += 1
count
fiveWeekends = ->
startYear = 1900
endYear = 2100
year = undefined
monthTotal = 0
yearsWithoutFiveWeekends = []
total = 0
year = startYear
while year <= endYear
monthTotal = checkMonths(year)
total += monthTotal
# extra credit
if monthTotal == 0
yearsWithoutFiveWeekends.push year
year += 1
console.log 'Total number of months: ' + total + ''
console.log ''
console.log yearsWithoutFiveWeekends + ''
console.log 'Years with no five-weekend months: ' + yearsWithoutFiveWeekends.length + ''
return
fiveWeekends()
1901 2
1902 7
1903 4
1904 0
1904 6
1905 11
1907 2
1908 4
1909 0
1909 9
1910 6
1911 11
1912 2
1913 7
1914 4
1915 0
1915 9
1916 11
1918 2
1919 7
1920 9
1921 6
1922 11
1924 7
..
Total number of months: 201
1900,1906,1917,1923,1928,1934,1945,1951,1956,1962,1973,1979,1984,1990,2001,2007,2012,2018,2029,2035,2040,2046,2057,2063,2068,2074,2085,2091,2096
Years with no five-weekend months: 29
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