RogerBW's Blog

The Weekly Challenge 256: Maximum Strings 18 February 2024

I’ve been doing the Weekly Challenges. The latest involved array analysis and string merging. (Note that this ends today.)

You are given an array of distinct words, `@words`.

Write a script to find the maximum pairs in the given array. The words `\$words[i]` and `\$words[j]` can be a pair if one is the reverse of the other.

The trick, of course, is that I only want to count each pair once. So I end up doing what feels like half the problem. In Raku:

``````sub maximumpairs(@a) {
``````

Set up pair count and cache.

``````    my \$n = 0;
my \$r = SetHash.new;
``````

For each word,

``````    for @a -> \$s {
``````

Get its reverse.

``````        my \$t = \$s.flip;
``````

If I've already seen the reverse, add one to the pair count.

``````        if (\$r{\$t}:exists) {
\$n++;
``````

Otherwise, store the word in the cache so that we'll spot its reverse later.

``````        } else {
\$r{\$s} = 1;
}
}
return \$n;
}
``````

This will be inconsistent if there are duplicates (e.g. [ba ab ab] would score 2, while [ab ab ba] would only score 1) but the problem doesn't specify how to treat those anyway.

You are given two strings, \$str1 and \$str2.

Write a script to merge the given strings by adding in alternative order starting with the first string. If a string is longer than the other then append the remaining at the end.

So it's the functionality which in many languages is called something like `zip`, but carrying on until both are exhausted rather than ending when the first one ends. I'm pretty sure Rust has an option for doing this automatically, but since I had to write the algorithm from scratch for some of the languages anyway, it was easier just to use it everywhere.

JavaScript:

``````function mergestrings(a, b) {
let out = "";
``````

Iterate over length of longest string.

``````    for (let i = 0; i < Math.max(a.length, b.length); i++) {
``````

If we haven't gone off the end of `a`, add the character from this place in `a`.

``````        if (i <= a.length - 1) {
out += a.substring(i, i+1);
}
``````

If we haven't gone off the end of `b`, add the character from this place in `b`.

``````        if (i <= b.length - 1) {
out += b.substring(i, i+1);
}
}
return out;
}
``````

(I'm doing that classic computing thing of treating 2 as a special case of 1 rather than as a special case of infinity. An `n`-way merge would put the strings in an array and cycle over it for each character.)

Full code on github.