Weekly Challenge: The last word is my buddy

Weekly Challenge 331

Each week Mohammad S. Anwar sends out The Weekly Challenge, a chance for all of us to come up with solutions to two weekly tasks. My solutions are written in Python first, and then converted to Perl. It's a great way for us all to practice some coding.

Challenge, My solutions

Task 1: Last Word

Task

You are given a string.

Write a script to find the length of last word in the given string.

My solution

Sometimes I over complicate things, and this might be one of those cases. We'll start with "what is a word?". Is isn't a word? Spoiler: it is, and it's not a one letter word.

What if the string is Perl Weekly Challenge!. In this case, the last word is nine letters long. So for this task, I use a regular expression to look for the last letter, and any combination of letters, hyphens and apostrophes before it. I then return the length of this string or zero if it doesn't exist.

def last_word(input_string: str) -> int:
    # Find the last word in the input string and return its length
    match = re.search(
        r"([a-z'-]*[a-z])[^a-z]*$",
        input_string.strip(),
        re.IGNORECASE
    )
    return len(match.group(1)) if match else 0

The Perl code is a little more straight forward.

sub main ($input_string) {
    my ($word) = $input_string =~ /([a-z'-]*[a-z])[^a-z]*$/i;
    say length($word // '');
}

Examples

$ ./ch-1.py "The Weekly Challenge"
9

$ ./ch-1.py "   Hello   World    "
5

$ ./ch-1.py "Let's begin the fun"
3

Task 2: Buddy Strings

Task

You are given two strings, source and target.

Write a script to find out if the given strings are Buddy Strings.

If swapping of a letter in one string make them same as the other then they are Buddy Strings.

My solution

For this task, I start by checking the strings are of the same length. If they are not, I return False.

I then use the combinations function from itertools to get all unique combinations of character positions (from 0 to one less than the length of the string), assigned to the variables i and j. As strings are immutable in Python, I convert the source string to a swapped list, transpose the characters at position i and j and check if this is the same a the target string. If it is, I return True. If the loop is exhausted, I return False.

def buddy_strings(source: str, target: str) -> bool:
    if len(source) != len(target):
        return False

    for i, j in combinations(range(len(source)), 2):
        swapped = list(source)
        swapped[i], swapped[j] = swapped[j], swapped[i]
        if ''.join(swapped) == target:
            return True

    return False

The Perl solution follows the same logic.

Examples

$ ./ch-2.py nice ncie
True

$ ./ch-2.py love love
False

$ ./ch-2.py feed feed
True