I’ve been doing the Weekly
Challenges. The
latest
involved date calculations and didn't involve array permutations.
(Note that this ends today.)
Task 1: Seize The Day
Given a year, a month, a weekday of month, and a day of week (1 (Mon) .. 7
(Sun)), print the day.
Rust's chrono
has a built-in for this. Har har. For languages that
don't:
- find weekday of first day of month
- find date of first desired weekday in month
- add 7 per subsequent week count
- check whether this is still in the same month
Raku:
sub seizetheday($year, $month, $weekcount, $dayofweek) {
Start at the first day of the month.
my $dt = Date.new(year => $year,
month => $month,
day => 1);
Get its weekday.
my $wd = $dt.day-of-week;
If that's not the weekday we want, move it forward to the first
one that is.
if ($wd != $dayofweek) {
$dt = $dt.later(days => ($dayofweek - $wd + 7) % 7);
}
Then if we want a later X-day, step forward by weeks.
if ($weekcount > 1) {
$dt = $dt.later(weeks => ($weekcount - 1));
}
If this modified date isn't in the same month and year as
the original, return zero.
if ($dt.month != $month || $dt.year != $year) {
return 0;
}
Return the day of the month.
return $dt.day;
}
Lua and PostScript don't have date libraries. Well, they do now. All
it really needs is a converter from y-m-d to Julian date, a converter
the other way—both of which you can readily build from the formulae on
the Wikipedia
page—and
a day-of-week calculator which is essentially Julian date mod 7 with
an offset.
Kotlin and Scala, both using the Java time
library, end up looking
almost identical.
Task 2: Maximise Greatness
You are given an array of integers.
Write a script to permute the give array such that you get the maximum
possible greatness.
To determine greatness, nums[i] < perm[i]
where 0 <= i < nums.length
The hardest part here is the description of the problem, because it
might lead one astray into actually trying permutations.
First, sort the list. Then have a pair of pointers, let's call them
lead
and lag
; start both at zero.
Run lead
over each list element. If the value there is greater than
the value at lag
, increment lag
by one. (Because the list is
sorted, this is the lowest possible lead
value that's greater than
the lag
value.)
The final value of lag
is the greatness.
In practice we aren't using C so lead
is implemented as an iterator
over the sorted list.
Ruby:
def maximisegreatness(a)
b = a.sort
g = 0
b.each do |c|
if c > b[g] then
g += 1
end
end
return g
end
Full code on
github.
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