New Year's Day clear solution for The Most Frequent Weekdays by gyahun_dash - python coding challenges

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New Year's Day solution in Clear category for The Most Frequent Weekdays by gyahun_dash

from calendar import day_name, isleap, weekday 

def most_frequent_days(year):
    days = (weekday(year, 1, 1 + i) for i in range(1 + isleap(year)))
    return [day_name[d] for d in sorted(days)]

March 22, 2016

Comments:

e-l on March 25, 2016, 9:31 a.m.
How does it work? You are taking weekdays for Jan 1 and Jan 2 for leap years. Why are they the most frequent days?

suic on March 25, 2016, 2:20 p.m.
Hi, see the discussion under <a href="http://www.checkio.org/mission/the-most-frequent-weekdays/publications/suic/python-3/only-the-first-and-the-last-week-matters/">my solution</a>.

gyahun_dash on March 25, 2016, 2:38 p.m.
There are 365 (or 366) days and 365 // 7 = 52 weeks in 1 year. The most frequent days are the remaining 365 % 7 = 1 day (or 366 % 7 = 2 days). So all we need to do is to check weekdays of Dec 31 (and 30). But these are equal to the first weekdays of the year, i.e. Jan 1 (and 2).

saharnasiri71 on April 11, 2016, 10:30 a.m.
I did not get how does weekday(year, 1, 1+i) return 52 weeks!

singhcoolguy on April 12, 2016, 7:02 a.m.
As the name implies, weekday(year, month, day) returns the weekday for the particular date. We don't require 52 weeks. We only need return weekday for Jan 1 ( and 2 if it's a leap year) and that's exactly what it does.

ljy95135 on April 18, 2016, 3:06 a.m.
Awsome! It's indeed right after pointing it out.

AlexanderKamenev on Oct. 4, 2016, 7:35 a.m.
Awsome!

pathFinder586 on Jan. 25, 2022, 10:20 p.m.
I feel like an idiot after seeing this solution. :-(

Master of the Time Stone

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