RogerBW's Blog

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

Task 1: Check Order

You are given an array of integers, @ints.

Write a script to re-arrange the given array in an increasing order and return the indices where it differs from the original array.

Algorithmically, this is straightforward: copy the list, sort the copy, compare the two and return the indices of differences. Raku:

sub checkorder(@a) {

Build the sorted list.

    my @b = @a.sort({$^a <=> $^b});
    my @out;

Compare each index.

    for @b.kv -> $i, $c {
        if (@a[$i] != $c) {
            @out.push($i);
        }
    }
    @out;
}

Task 2: Find Anagrams

You are given a list of words, @words.

Write a script to find any two consecutive words and if they are anagrams, drop the first word and keep the second. You continue this until there is no more anagrams in the given list and return the count of final list.

For the anagram test, I chose to sort the characters o each word and test that list for string equality.

Therefore "drop the first word and keep the second" becomes a meaningless constraint, since they're identical. All I need to do is examine each consecutive pair of words and add one to the count if they're different. (A more usual formulation of this kind of problem would ask for the number of different words overall, in which case I'd stuff them in a set and return the length.) Kotlin:

fun findanagrams(a: List<String>): Int {

Populate the working list with sorted strings.

    var b = ArrayList<String>()
    for (s in a) {
        b.add(s.toList().sorted().joinToString(""))
    }

Count the number of times an adjacent pair of words differs (adding 1 because we don't compare the first word with a notional empty string at the start of the list).

    var out = 1
    for (s in b.windowed(size = 2)) {
        if (s[0] != s[1]) {
            out += 1
        }
    }
    return out
}

fun main() {

    if (findanagrams(listOf("acca", "dog", "god", "perl", "repl")) == 3) {
        print("Pass")
    } else {
        print("Fail")
    }
    print(" ")
    if (findanagrams(listOf("abba", "baba", "aabb", "ab", "ab")) == 2) {
        print("Pass")
    } else {
        print("Fail")
    }
    println("")

}

Full code on github.