codewar challenge javascript

Question

I hope everyone is having a good day.

This is my first ever post on Stackoverflow!

I have just completed the javascript course on codeacademy and have read a couple of books on it too. Now I am on codewars. I would classify myself as a beginner in Javascript.

I find myslef a little stuck on a challenge, please can someone shed some light on what I may be doing wrong? Many thanks in advance!

Here is the instructions:

Check to see if a string has the same amount of 'x's and 'o's. The method must return a boolean and be case insensitive. The string can contains any char.

And here is my code:

function XO(str) {
    var x = [];
    var o = [];

    for (var i = 0; i <= str.length; i++) {
        if (str(i).toLowerCase === "x") {
            x.push(i);
        } else if (str(i).toLowerCase === "o") {
            o.push(i);
        }

        if (x.length === o.length) {
            return true;
        } else {
            return false;
        }
    }
}

Move your length check out of the for loop

ASDFGerte
– ASDFGerte

2017-08-13 14:34:17 +00:00
Commented Aug 13, 2017 at 14:34 — ASDFGerte
– ASDFGerte, Commented Aug 13, 2017 at 14:34
your codes are not good format yet

raksa
– raksa

2017-08-13 14:36:22 +00:00
Commented Aug 13, 2017 at 14:36 — raksa
– raksa, Commented Aug 13, 2017 at 14:36
why not use inbuilt string functions ?

marvel308
– marvel308

2017-08-13 14:40:12 +00:00
Commented Aug 13, 2017 at 14:40 — marvel308
– marvel308, Commented Aug 13, 2017 at 14:40

raksa · Accepted Answer · 2017-08-13 14:56:28Z

1

i corrected mistakes and use code comment to explain

function XO(str) {
    var x = [];
    var o = [];
    for (var i = 0; i < str.length; i++) { // i must be lower than length
        // str[i] or str.charAt(i), not str(i)
        if (str[i].toLowerCase() === 'x') { // toLowerCase is function, it must be called with toLowerCase()
            x.push(str[i]); // should push character
        } else if (str[i].toLowerCase() === 'o') {
            o.push(str[i]);
        }
    }
    // return statement must be located at the end
    if (x.length == o.length) {
        return true;
    } else {
        return false;
    }
}
console.log(XO('xo'));
console.log(XO('xxo'));
console.log(XO('xoX'));
console.log(XO('xoOX'));

answered Aug 13, 2017 at 14:56

raksa

9488 silver badges19 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

CoderAz Over a year ago

Thank you very much, and thank you everyone else who responded. I just wanted to also ask, what is the difference if i used [i] instead of (i)...I know [i] works in a way to go through the index of arrays but I get a bit confused when people use it in the (i) method. I know this may be a sily question to many but as mentioned I am just starting out in javascript. Thanks guys

Ashvin777 Over a year ago

Parenthesis are used for method calling. Like str[i].toLowerCase(). Here toLowerCase is a method. Angle brackets [] are used for accessing or changing the values in JSON or Array data objects. Both are totally different things.

Djaouad · Accepted Answer · 2020-05-22 23:53:35Z

1

function XO(str) {
    var x = 0, // numbers are better
        o = 0;

    for (var i = 0; i < str.length; i++) { // changed from '<=' to '<'
        if (str[i].toLowerCase() === "x") {
            x++;
        } else if (str[i].toLowerCase() === "o") {
            o++;
        }
    }
    return x === o;
}

edited May 22, 2020 at 23:53

answered Aug 13, 2017 at 14:35

Djaouad

22.8k4 gold badges37 silver badges57 bronze badges

Comments

Siphalor · Accepted Answer · 2017-08-13 14:33:47Z

0

str.match(/x/g).length==str.match(/o/g).length

answered Aug 13, 2017 at 14:33

Siphalor

7246 silver badges16 bronze badges

Comments

marvel308 · Accepted Answer · 2017-08-13 14:34:31Z

0

function checkIfOequalsX(str){
  return str.match(/x/g).length==str.match(/o/g).length
}

console.log(checkIfOequalsX('xxooxo'));
console.log(checkIfOequalsX('xxooxoo'));

you can do it with

str.match(/x/g).length==str.match(/o/g).length

answered Aug 13, 2017 at 14:34

marvel308

10.5k3 gold badges24 silver badges32 bronze badges

Comments

Ashvin777 · Accepted Answer · 2017-08-14 05:05:29Z

0

The third if else will never be executed because for a string there will be always a value.

If you want to return the count then the check of length should be performed after the for loop.

var xCount = 0; var oCount = 0;
for (var i = 0; i < str.length; i++) {
  if (str[i].toLowerCase() === "x") {
    xCount++;
  } else if (str[i].toLowerCase() === "o") {
    oCount++;
  }
}

return xCount === oCount;

About another solutions containing the check based on str.match method, the complexity of using that piece of code is twice compared to above because the str.match loop is performed twice to match both the strings.

edited Aug 14, 2017 at 5:05

answered Aug 13, 2017 at 14:33

Ashvin777

1,48213 silver badges19 bronze badges

6 Comments

James Hamann Over a year ago

This is the best answer since you only have to loop once.

marvel308 Over a year ago

the complexity is the same, if you loop once or twice. The number of instructions executed matters

ASDFGerte Over a year ago

As another answer already explained, there is another problem with <= str.length.

James Hamann Over a year ago

Sure but isn't the O time O(n) vs O(2n)?

Djaouad Over a year ago

You need to use toLowerCase() not toLowerCase, because the latter is a function code, while the former is the value that the function returns, in other words, str(i).toLowerCase === "x" is true for any character str(i).

|

Paul Roub · Accepted Answer · 2020-05-22 22:44:40Z

0

function XO(str) {
  let strn = str.toLowerCase();
  let countX = []; 
  let countO = []; 
  for(let i=0; i<strn.length; i++) {
    if(strn[i] == 'x') {
      countX.push(strn[i])
    } else if(strn[i] == 'o') {
      countO.push(strn[i])
    }
  }

  if(countX.length == countO.length){
    return true
  } else if(countX.length !== countO.length) {
    return false
  }
}

edited May 22, 2020 at 22:44

Paul Roub

36.5k27 gold badges88 silver badges95 bronze badges

answered May 22, 2020 at 22:39

Max Zhukov

1

Comments

namhd · Accepted Answer · 2020-12-16 14:58:57Z

0

You can use the Regex for finding those characters:

function XO(str) {
 return str.match(/o/ig).length === str.match(/x/ig).length;
}

answered Dec 16, 2020 at 14:58

namhd

3372 silver badges12 bronze badges

Comments

Minal Chauhan · Accepted Answer · 2022-01-03 05:16:54Z

0

function XO(str) {
   let letraO = 0
   let letraX = 0
   const myArray = str.toLowerCase();
   for (i=0; i<myArray.length; i++){
      myArray[i] === 'o'? letraO++ : letraX ++
   }
   return letraO===letraX? true : false    
}

edited Jan 3, 2022 at 5:16

Minal Chauhan

6,16812 gold badges24 silver badges41 bronze badges

answered Jan 3, 2022 at 2:25

Eleonora

1

Comments

A100D · Accepted Answer · 2022-09-15 02:31:42Z

0

function XO(str) {

    // make the string lowercase because we are case insensitive
    str = str.toLowerCase();

    // put the string into an array
    var arrayOfCharacters = str.split("");

    //count the x's
    var countX = arrayOfCharacters.reduce( function( n, val ) {
        return n + (val === 'x');
      }, 0);
      
    // count the o's
    var countO = arrayOfCharacters.reduce( function( n, val ) {
        return n + (val === 'o');
      }, 0);
    
    // do these numbers match? if so return true and if not return false
    if ( countX == countO ) {
      return true;
    } else {
      return false;
    }
}

answered Sep 15, 2022 at 2:31

A100D

116 bronze badges

Collectives™ on Stack Overflow

codewar challenge javascript

9 Answers 9

2 Comments

Comments

Comments

Comments

6 Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

9 Answers 9

2 Comments

Comments

Comments

Comments

6 Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related