Daily Challenge #29 - Xs and Os

I can't do anything useful with it yet, but it's fun to solve coding katas :D.

Yes. Although mine looks a bit uglier because I ran into an issue: if the string doesn't have Xs or Os, the result of match is null so it fails when trying to do .length (test checkXO("zpzpzpp") on your code).

I had to add the fallback into empty array for those cases. But I feel like there has to be a cleaner way of doing the same thing.

Hi and no problem changing the code

The following is indeed not needed:

(xc == oc ? true : false);

The loop tests strings from both sides to be more performant. Try doing benchmarks of both versions and post the results. I don't have time right now but might do it in the evening and show you the difference. Or, I might just embarrass myself lol.

Cheers

I wrote this too fast without even thinking too much, so once again, I appreciate your comment. Anyway, I know that inlining helps, but speed wasn't on my priority list for this challenge. :)

Here are my results:

Your code: ~550 nsec
My code: ~1140 nsec

Thumbs up for faster code.

P.S.
I suspected that strlen might be the culprit, and I was right.

If I change the code like this, I get results similar to yours, around 500 nsec more/less

static bool count_xo(const char* d, size_t l){
    // null pointer
    if(!d) return true;
    // no data
    if(!l) return true;
    // res counters
    unsigned xc = 0, oc = 0;
    // mid point
    unsigned mp = l / 2, rm = l % 2;
    // check for 88 (X) and 79 (O)
    // O(N/2)
    for(unsigned i = 0, j = l - 1; i < mp; i++, j--){
        // X counter
        xc += check_x(d[i]) + check_x(d[j]);
        // O counter
        oc += check_o(d[i]) + check_o(d[j]);
    }
    // remainder
    if(rm){
        // X counter
        xc += check_x(d[mp + 1]);
        // O counter
        oc += check_o(d[mp + 1]);
    }

    // res
    return xc == oc;
}

Well that's C, just how it is, you either hate it or you love it 😁. I missed a comma there; what I wanted to say was: guys, please don't kill me, this is C and it's gonna get ugly.

Sure, but this one is pretty easy to demystify, luckily: tr/set1/set2/ just replaces characters in set1 with those in set2 (positionally), and returns the number of characters replaced/deleted. So in this code, he just compares the number of X's replaced with the number of O's replaced, and includes the lower-case variants too.

The local($_) is just there to let him make the code more brief. $_ is the "default input/pattern-searching space", so its like the default argument that tr will search. But it's also a global, so this lets you make a local copy and set it to be the first argument to the function.

If he didn't do that, it might look like this:

sub xo { my $xos = shift; ($xos =~ tr/xX//) == ($xos =~ tr/oO//) }

So just a bit more verbose.

It's working for me in chrome just now. Try wrapping the output line in a console.log if you're outside a REPL

In the reduce acc is the accumulator, keeping a sum of values as they come up. The reduce returns the sum at the end.
Each c is a character from the string (...str is an array of the chars from str)
c == 'x' || c == 'X' is obv a boolean. The code uses the fact that in JS Number(true) == 1 and Number(false) == 0, so it's a shorter way of saying:

XO = str => 
  [...str].reduce(
    (acc, c) => acc + (c == 'x' || c == 'X' ? 1 : 0) - (c == 'o' || c == 'O' ? 1 : 0),
    0
  ) == 0;

So each x adds one to the sum, and each o subtracts one. If the sum is zero there are the same number of x's as o's

Thank you Jacob! I have the same opinion! 👍

It turns out that you could avoid the #downcase with string.count("xX") == string.count("oO")

I don't know whether the performance would be significantly different, though it would avoid creating a couple of extra string instances.

There are some other interesting variations for #count, such as using it to count lower case letters with "abGvf".count("a-z")

Very nice! I love your choice of method name 😂 And it's great that you went back to refactor once you realized there were optimizations you could make. No matter what "stakeholders" say about features driving revenue, any developer worth their salt knows that a clean codebase drives features. 👍👍🙌