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# 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 yo+challenge@dev.to with your suggestions!

## Discussion (60)

Avalander

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.

Damir Franusic • Edited on

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 • Edited on

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:

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 • Edited on

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

Damir Franusic • Edited on

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

Josh

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

Damir Franusic • Edited on

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 • Edited on

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 • Edited on

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 :)

Alvaro Montoro • Edited on

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.)

willsmart • Edited on

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 • Edited on

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

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! 👍

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!

Shannon Crabill • Edited on

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 • Edited on

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!

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 • Edited on

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.

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);
}

}
Josh • Edited on
const exsAndOhs = // "🎙 they haunt me…" (https://www.youtube.com/watch?v=0uLI6BnVh6w)
(string) => string
.toLowerCase()
.split("")
.reduce((count, letter) =>
/x/.test(letter) ? count += 1
: /o/.test(letter) ? count -= 1
: count
, 0) === 0;

exsAndOhs("xoxo") // => true
exsAndOhs("like ghosts, they want me") // => false
exsAndOhs("to make them all, they won't let go") // => false
exsAndOhs("ex's and ohs") // => true

TERNARY CHAINS?? Yeah. I'm a loner, Dottie. A rebel…

Josh • Edited on

Now in elixir because hey all the cool kids are using not-Javascript here.

defmodule ExsAndOhs do

def try_to_get_over_them(string) do
string
|> String.downcase
|> test(0) # The `Enum` module's for chumps.
end

defp test("", count), do: count == 0  # whole string is traversed
defp test("x" <> string, count), do:  # next grapheme is an "x"
test(string, count + 1)
defp test("o" <> string, count), do:  # next grapheme is an "o"
test(string, count - 1)
defp test(string, count), do:         # String.next_grapheme/1 is a tuple 😃
string
|> String.next_grapheme
|> elem(1)
|> test(count)

end
SavagePixie

Hmmm... Something like this might work in JavaScript

const checkXO = str => str.match(/x/gi).length == str.match(/o/gi).length
Craig McIlwrath

import Data.Char (toLower)

xo :: String -> Bool
xo xs = let o's = count 'o' str
x's = count 'x' str
str = map toLower xs
count c = foldl (\sum char -> if c == char then sum + 1 else sum) 0
in o's == x's
Mario See

A little Python here.

def count_xo(s):
count_x, count_o = 0, 0
for c in s:
if c.lower() == "x":
count_x += 1
if c.lower() == "o":
count_o += 1
return count_x == count_o

This method uses a for loop to go through each character. It converts each character to lower case for a case insensitive comparison with "x" and "o" and increments the respective counters. The return line checks for count equality to return a boolean.

The time complexity in big O notation is O(n) or linear, where n is the size of input string s, and the space complexity is O(1) or constant since it just uses two variables to keep track of counts.

Colin Bounouar

Is there a reason why you prefer a for loop over a single call to lower on the str, and then returning lowered.count('x') == lowered.count('o') ?

Jacob Evans • Edited on

I broke out the work a little more, I tried to be a tad clever and utilize switch but wasn't able to get it to work quite the way I wanted.

Hopefully, the slightly more vanilla JS will be useful in some way.

function XO(s) {
const arr = [...s];
let countX = 0;
let countO = 0;
arr.map((ele, i) => {
if (ele === `x`) {
countX++;
delete arr[i];
}
if (ele === `X`) {
countX++;
delete arr[i];
}
if (ele === `o`) {
countO++;
delete arr[i];
}
if (ele === `O`) {
countO++;
delete arr[i];
} else {
delete arr[i];
}
});
const compare = countO === countX;
return compare;
}

console.log(XO(`xxxoooxxxoooxxxoooxoxoppppppppppppxoooxx`)); // true
// console.log(XO(`ooxx`)); // true
// console.log(XO(`xooxx`)); // false
// console.log(XO(`zpzpzpp`)); // true
// console.log(XO(`zzoo`)); // false
// console.log(XO(`zzoOOOoo`));
Jacob Evans

I was removing elements with plans of creating two for loops going in opposite directions to speed up the process... but that was for fun and it lost interest in it.

Oleksii Filonenko

Elixir:

defmodule XO do
def same(string),
do: count(string, "x") == count(string, "o")

defp count(string, letter) do
string
|> String.graphemes()
|> Enum.filter(&(&1 == letter))
|> Enum.count()
end
end
Philip Hallstrom

Ruby

def xo(str)
tally = 0
str.each_char do |c|
case c.upcase
when "X" then tally += 1
when "O" then tally -= 1
end
end
tally.zero?
end

Shorter solutions, but this only loops through the string once which might be nice for enormous strings.

Rong Sen Ng • Edited on

This can be done in linear time and linear space in terms of complexity by keeping track of the number of letter in a Map object as you traverse the string. Do note that the letter should be case insensitive which that actually gives us a hint that we can make each of the character to be always lowercased before using it as a cache key.

Pseudocode:

a = 'xoxo'
m = Map()
loop for each n in a {
l = n.toLowerCase()

m.set(l, 1) if not m.has(l)
else m.set(l, 1 + m.get(l))
}

return True if not m.has('x') and not m.has('o')

return m.get('x') == m.get('o');

raistlin-hess

Javascript with no loops:

function XO(str) {
let str_arr = str.toLowerCase()
.replace(/[^xo]/g, '')
.split('')    //Convert String into an Array to allow using Array.sort()
.sort();

//All of the o's will be placed before the x's thanks to Array.sort()
//Add 1 to account for 0-indexing and to turn -1 into 0 when there are no o's
let oCount = str_arr.lastIndexOf('o') + 1;
let xCount = str_arr.length - oCount;

return !(xCount - oCount);
}

//Run the test cases
const testCases = [
"ooxx",
"xooxx",
"ooxXm",
"zpzpzpp",
"zzoo"
];
for(str of testCases) {
let result = XO(str);
console.log(`"\${str}" => \${result}`);
}

Python goes there :

def XO(input):
x_count = 0
o_count = 0
for c in input:
if c.lower() == 'x':
x_count += 1
elif c.lower() == 'o':
o_count += 1
return bool(x_count == o_count)

[V2] - One liner :

XO = lambda input : bool(input.lower().count('x') == input.lower().count('o'))

Mat-R-Such

Python:

def xo(txt):
return  True if txt.lower().count('x') == txt.lower().count('o') else False
K.V.Harish

My solution in js

const XO = (str) => {
const countX = (`\${str}`.match(/x/gi) || '').length,
countO = (`\${str}`.match(/o/gi) || '').length;
return countX === countO;
};

Jess

Clojure

(defn xo [s]
(def freq (frequencies (seq s)))
(= (get freq \x) (get freq \o)))
Jess

Scala

def xo(s: String): Boolean = {
s.count(_ == 'x') == s.count(_ == 'o')
}