The Weekly Challenge 341

The Strategy: Store broken keys in a hash set for O(1) lookup. Split the string into words and for each word check every character (case-insensitive) against the broken set. Count words where no character matches a broken key.

sub broken_keyboard {
    my ( $str, @broken_keys ) = @_;
    my %broken = map { $_ => 1 } @broken_keys;

    my @words = split /\s+/, $str;
    my $count = 0;

    foreach my $word (@words) {
        my $can_type = 1;
        foreach my $char ( split //, lc($word) ) {
            if ( exists $broken{$char} ) {
                $can_type = 0;
                last;
            }
        }
        $count++ if $can_type;
    }

    return $count;
}

def broken_keyboard(text: str, *broken_keys: str) -> int:
    """Count words that can be typed without using any broken key."""
    broken = set(broken_keys)

    def can_type(word: str) -> bool:
        return all(ch.lower() not in broken for ch in word)

    return sum(1 for word in text.split() if can_type(word))

The Strategy: Find the index of the first occurrence of the target character. If found, extract the prefix (up to and including the character), reverse it, and concatenate with the remainder. If not found, return the string as-is.

sub reverse_prefix {
    my ( $str, $char ) = @_;
    my $pos = index( $str, $char );

    return $str if $pos == -1;

    my $prefix = substr( $str, 0, $pos + 1 );
    return reverse($prefix) . substr( $str, $pos + 1 );
}

def reverse_prefix(text: str, char: str) -> str:
    """Reverse the prefix up to and including the first occurrence of char."""
    pos = text.find(char)
    if pos == -1:
        return text

    prefix = text[:pos + 1]
    return prefix[::-1] + text[pos + 1:]