Daily Challenge #29 - Xs and Os

Can you check to see if a string has the same amount of 'x's and 'o's?

Examples input/output:

XO("ooxx") => true
XO("xooxx") => false
XO("ooxXm") => true
XO("zpzpzpp") => true // when no 'x' and 'o' is present should return true
XO("zzoo") => false

Note: The method must return a boolean and be case insensitive. The string can contain any character

---

This challenge comes from user joh_pot. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge for a future post? Email [e-mail] with your suggestions!

Comments

[Avalander]

Haskell again

import Data.Char (toLower)

compare_xos :: String -> Bool
compare_xos = same . foldl kevin (0, 0)
  where
    kevin (x, o) c
      | toLower c == 'x' = (x + 1, o)
      | toLower c == 'o' = (x, o + 1)
      | otherwise        = (x, o)
    same (x, o) = x == o

[Josh]

dammit kevin

[Jacob Evans]

Just being able to use Haskell impresses me lol

[Avalander]

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

[Alvaro Montoro]

JavaScript

const xo = s => (s.match(/x/gi) || []).length == (s.match(/o/gi) || []).length;

A solution using regular expressions in JavaScript. Live demo on CodePen.

[SavagePixie]

Oh look, you had already come up with the same idea as I did!

[Alvaro Montoro]

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.

[SavagePixie]

Yeah, I found the same issue when I tried to solve it in Codewars. So I had to do the same. Looking through other people's solutions, there's one that I find interesting that uses split() instead of a regular expression. Something like this:

const xo = str => str.toLowerCase().split('x').length == str.toLowerCase().split('y').length

[Alvaro Montoro]

The split approach is interesting. I was playing with different options, but not with this. Also, I found that using toLowerCase() made the code considerably slower, it works considerably faster if you use a regular expression in the split:

const xo6 = str => str.split(/x/i).length == str.split(/o/i).length;

[SavagePixie]

I think this is my favourite solution. I hadn't checked its speed, but I thought that toLowerCase() probably didn't make it any smoother than returning an empty array if there are no matches.

[Alvaro Montoro]

I checked different solutions with jsperf and this was the second fastest by a little (but results may vary.)

[Damir Franusic]

C 😱 don't kill me please

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

// bool as string
#define BOOL_STR(b) b ? "true" : "false"

// X/x check
static bool check_x(char d) {
    if (d == 88 || d == 120) return true;
    return false;
}

// O/o check
static bool check_o(char d){
    if (d == 79 || d == 111) return true;
    return false;
}

static bool count_xo(const char* d){
    // null pointer
    if(!d) return true;
    // get size
    size_t l = strlen(d);
    // 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 ? true : false);
}

int main(void) { 
    // test strings
    const char* str_00 = "ooxx";
    const char* str_01 = "xooxx";
    const char* str_02 = "ooxXm";
    const char* str_03 = "zpzpzpp";
    const char* str_04 = "zzoo";
    // results
    printf("%s: %s\n", str_00, BOOL_STR(count_xo(str_00)));
    printf("%s: %s\n", str_01, BOOL_STR(count_xo(str_01)));
    printf("%s: %s\n", str_02, BOOL_STR(count_xo(str_02)));
    printf("%s: %s\n", str_03, BOOL_STR(count_xo(str_03)));
    printf("%s: %s\n", str_04, BOOL_STR(count_xo(str_04)));
    // no error
    return 0; 
}

Results:

ooxx: true
xooxx: false
ooxXm: true
zpzpzpp: true
zzoo: false

[Olivier “Ölbaum” Scherler]

Is it allowed to modify someone else’s solution? I think you made it a bit more complicated than necessary. Anyone please correct me if I’m wrong:

• C strings end with the null character '\0', so you can skip getting the string length and do a while loop that checks for the null character (strlen does just that anyway);

• I didn’t really understand why you’re walking the string from the start and the middle, instead of from beginning to end, so I changed it;

• since we then only test for x and o once, I could inline the tests;

• xc == oc and d == 88 || d == 120 are already boolean expressions, so you don’t need the ternary operator ... ? true : false or an if(...) return true; else return false;;

• since I never used assertions in C, I took the opportunity to try them for the test cases (I intentionally wrote assert(count_xo("xooxx") == false) in the tests instead of assert(! count_xo("xooxx")) for consistency and readability.
  

#include <stdio.h>
#include <stdbool.h>

// bool as string
#define BOOL_STR(b) b ? "true" : "false"

static bool count_xo(const char* d){
    // null pointer
    if(!d) return true;
    // counters
    unsigned xc = 0, oc = 0, i = -1;
    while(d[++i] != '\0'){
        // X counter
        if( d[i] == 88 || d[i] == 120 ) xc++;
        // O counter
        if( d[i] == 79 || d[i] == 111 ) oc++;
    }
    // res
    return xc == oc;
}

int main(void) { 
    // test strings
    assert(count_xo("ooxx") == true);
    assert(count_xo("xooxx") == false);
    assert(count_xo("ooxXm") == true);
    assert(count_xo("zpzpzpp") == true);
    assert(count_xo("zzoo") == false);
    assert(count_xo("xoffxffo") == true);

    return 0; 
}

[Damir Franusic]

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

[Damir Franusic]

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;
}

[Olivier “Ölbaum” Scherler]

Nice. I didn’t inline for speed, but for compactness, so that people don’t say C is too complicated, so verbose, etc. :-)

[Damir Franusic]

C is my baby, just started embedded C on SBCs, mostly arm 32bit. And people will always say that 🤣👍

[Josh]

I feel like the C might've already done you in? dang, that's some verbose low-level lang 😶

[Damir Franusic]

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.

[E. Choroba]

Very easy (and rather fast) in Perl:

sub xo { local ($_) = @_; tr/xX// == tr/oO// }

It uses the transliteration operator which returns the number of matching characters in scalar context, and numeric comparison imposes scalar context.

The tests:

use Test::More;

ok   xo('ooxx');
ok ! xo('xooxx');
ok   xo("ooxXm");
ok   xo("zpzpzpp");
ok ! xo("zzoo");

done_testing();

[Damir Franusic]

Perl sorcery 😁. This syntax always turns me inside out hehe. Nice job 👍

[Yozen Hernandez]

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.

[Alvaro Montoro]

Thanks for a great explanation :)

[willsmart]

How about a reduce to keep it in one loop? (JS)

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

// Check it
["ooxx","xooxx","ooxXm","zpzpzpp","zzoo"].map(s =>
  `${s} => ${XO(s)}`
).join('\n');
/* ^
"ooxx => true
xooxx => false
ooxXm => true
zpzpzpp => true
zzoo => false"
*/

[Josh]

Oh, nice! I wouldn't have thought to use the array spread syntax to get the array of characters. 🎩✨

[nishu1343]

Bro, nice solution .cud u explain me what is going inside the reduce. And im not able to reproduce it.

[willsmart]

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

[nishu1343]

Cool.its working😃

[Rafael Pereira]

Javascript (pretty readable):

const makeArrayOfGivenChar = (text, givenChar) => text.split('').filter(c => c.toLowerCase() === givenChar.toLowerCase())

function XO(str) {
  const x = makeArrayOfGivenChar(str, 'x')
  const o = makeArrayOfGivenChar(str, 'o')
  return x.length === o.length
}

[Jacob Evans]

Well done! I like a one-liner and clever answers as much as the next person HOWEVER I also love seeing elegant, more readable, and even sometimes simplified (broken down more step by step) solutions!

Awesome job!

[Rafael Pereira]

Thank you Jacob! I have the same opinion! 👍

[Shannon Crabill]

Not a pretty solution, but it seems to work in Ruby

def gossip_girl(string)
  new_string = string.downcase

  if new_string.include?("x") == false && new_string.include?("o") == false
    true
  elsif new_string.include?("x") == true && new_string.include?("o") == true
    x_count = new_string.count("x")
    o_count = new_string.count("o")

    if x_count == o_count
      true
    else
      false
    end
  else
    false
  end
end

Returns

gossip_girl("ooxx") => true 
gossip_girl("xooxx") => false 
gossip_girl("ooxXm") => true 
gossip_girl("zpzpzpp") => true 
gossip_girl("zzoo") => false 

[Shannon Crabill]

WAIT. Here's a much cleaner solution in Ruby.

def gossip_girl(string)
  string.downcase.count("x") == string.downcase.count("o")
end

Which returns

gossip_girl("ooxx") => true 
gossip_girl("xooxx") => false 
gossip_girl("ooxXm") => true 
gossip_girl("zpzpzpp") => true 
gossip_girl("zzoo") => false 

This is why you refactor!

[MetaDave 🇪🇺]

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")

[Josh]

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. 👍👍🙌

[Shannon Crabill]

Thanks. I can't think of X's and O's without Gossip Girl.

Clean codebases are so much easier to work with. Not to mention, cost and performance savings.

[La blatte]

How about some ugly oneliner?

XO=p=>!(new Set(p.toLowerCase().split``.reduce((l,i)=>[l[0]+(i=='x'),l[1]+(i=='o')],[0,0])).size-1)

Which produces the following output:

XO("ooxx"); // true
XO("xooxx"); // false
XO("ooxXm"); // true
XO("zpzpzpp"); // true
XO("zzoo"); // false

It basically works by lowercasing everything and building an array with the number of Xs and Os. After that, it builds a Set with it, and check if the Set size is 1 or 2!

[Mohamed ELIDRISSI]

This is my solution, I'm still learning javascript so nothing fancy

function XO(string) {
  const strArr = string.toLowerCase().split('');
  let os = 0, xs = 0;
  strArr.forEach(word => {
    switch (word) {
      case 'o': 
        os++;
        break;
      case 'x': 
        xs++;
        break;
    }
  });
  return os === xs;
}

[Peter]

Rust

pub fn xo (value: &str) -> bool {
    let value = value.to_lowercase();
    let count_x = value.matches("x").count();
    let count_o = value.matches("o").count();
    count_x == count_o
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_1() {
        assert_eq!(xo("ooxx"), true);
    }
    #[test]
    fn test_2() {
        assert_eq!(xo("xooxx"), false);
    }
    #[test]
    fn test_3() {
        assert_eq!(xo("ooxXm"), true);
    }
    #[test]
    fn test_4() {
        assert_eq!(xo("zzoo"), false);
    }
    #[test]
    fn test_5() {
        assert_eq!(xo("zpzpzpp"), true);
    }

}

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