Tag: Perl

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #2 this week is described as follows:

You are given a list, @L, of three or more random integers between 1 and 50. A Noble Integer is an integer N in @L, such that there are exactly N integers greater than N in @L. Output any Noble Integer found in @L, or an empty list if none were found.

An interesting question is whether or not there can be multiple Noble Integers in a list.

For example, suppose we have list of 4 integers [2, 6, 1, 3].

Here we have 2 in the above list, known as Noble Integer, since there are exactly 2 integers in the list i.e. 3 and 6, which are greater than 2.

Therefore the script would print 2.

While Mohammad gave me credit for submitting this problem, I only suggested some wording changes right before it was published, so I didn’t have any sort of advantage going in.

The algorithm I came up with for finding Noble Integers is fairly simple and seems obvious: simply sort the array, and then for each array index, $i, @L.end - $i is the number of elements that come after. @L.end in Raku is $#L in Perl: the last index in the array.

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #1 this week asks us to merge integer ranges. Here is the task description:

Write a script to merge the given intervals where ever possible.

[2,7], [3,9], [10,12], [15,19], [18,22]

The script should merge [2, 7] and [3, 9] together to return [2, 9]. Similarly it should also merge [15, 19] and [18, 22] together to return [15, 22]. The final result should be something like below:

[2, 9], [10, 12], [15, 22]

Of course, we could brute force it by comparing every interval with every other interval for an O(n²) solution. But there is a relatively straightforward algorithm to solve this in O(n log n) time.

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #2 this week has us build an LRU cache, which is a cache of fixed capacity, whereby when it is full but a new item needs to be added, the item that was least-recently accessed is evicted from the cache.

For example, if a cache of capacity = 3 contains [a b c] (where a was accessed most recently), elements are evicted off the right side of the list. Adding element d, the list would contain [d a b], for example.

Of course, [d a b] are just the keys of the items in the cache. We’ll take an arbitrary scalar (which could itself be a reference) as a value.

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #1 this week is as follows:

Write a script to accept a positive number as command line argument and print the smallest multiple of the given number consists of digits 0 and 1.

Brute Force

It’s easy enough to brute force this problem, by checking every multiple in order:

$_ += $_[0] while /[^10]/;

$_ now contains the answer. Unfortunately, some numbers—particularly multiples of 9—require extremely large multiples. For example, 99 × 1122334455667789 = 111111111111111111 takes several minutes to discover via this method.

We can do better, though.

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #1 this week is described as follows:

There are 50 people standing in a circle in position 1 to 50. The person standing at position 1 has a sword. He kills the next person i.e. standing at position 2 and pass on the sword to the immediate next i.e. person standing at position 3. Now the person at position 3 does the same and it goes on until only one survives.

Write a script to find out the survivor.

This is a well known theoretical exercise in computer science, by the name of the Josepheus problem, based on the story of Flavius Josephus, a Jewish historian in the first century.

Algorithm

To me, the most natural way to solve this problem is with a circular linked list. As a very quick review, a linked list is a list of items where each item also contains a pointer, reference, index, whatever to the next item in the list. A single element is usually drawn like this:

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #2 this week asks us to (and I quote): print all Palindrome Dates between 2000 and 2999. The format of date is mmddyyyy. For example, the first one was on October 2, 2001 as it is represented as 10022001.

It’s pretty easy to avoid using any sort of date library with a couple of key observations about the problem domain. First, we’ll split it up into month (mm), day (dd), century (cc), and 2-digit year (yy). Thus our “baseline” date format is mm dd cc yy.

If we use R[xx] as shorthand for “string reverse of xx“, we can rewrite the date in a couple of different ways:

  mm    dd    cc    yy     # Original
R[yy] R[cc]   cc    yy     # Start with year
  mm    dd  R[dd] R[mm]    # Start with mm/dd
R[yy]   dd  R[dd]   yy     # Start with yy and dd

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task 2 this week has us print the first 20 “gapful numbers,” as described by OEIS sequence A108343. Gapful numbers are numbers greater than 99 that are evenly divisible by their first and last digits combined. For example, 132 is a gapful number because 132 ÷ 12 = 11.

This is certainly the easier of the two tasks, as both Perl and Raku have convenient ways to index and concatenate string fragments.