Why indexOf in javascript not working?

Question

I'm not sure what I'm doing wrong here. The first instance that I use indexOf it works perfectly fine, but when I use it the second time it's not returning the result that I'm expecting.

function mutation(arr) {
    //return arr;
    res = "";
    for (var x=0; x<arr[1].split("").length; x++) {
        if (arr[0].indexOf(arr[1].split("")[x]) !== -1) {
            res += "t";
        } else {
            res += "f";
        }
    }
    // res = ttt
    if (res.indexOf("f") !== -1) {
        return true;
    } else {
        return false;
    }
}

mutation(["hello", "hey"]);
// this returns true instead of false
mutation(["floor", "loo"]); 
// returns false instead of true

mutation should return false if an element from arr[1] is not present in arr[0] else return true.

I just pasted your code into the JS console and I got true as well — csga5000
– csga5000, Commented Nov 22, 2015 at 5:28
If you are using IE8 and olders, .indexOf() doesn't be supported. Reference link: stackoverflow.com/questions/3629183/… — user4090029
– user4090029, Commented Nov 22, 2015 at 5:32
@splucena check out my edited answer. The problems with your code are outlined there. You had an off by 1 error and a negated indexOf check. — Leah Zorychta
– Leah Zorychta, Commented Nov 22, 2015 at 5:41

Leah Zorychta · Accepted Answer · 2015-11-22 05:39:00Z

3

your code isn't working because when you say:

res.indexOf("f") != -1

this means: "I found an f", but you're treating it as if it means "I did not find an f".

In your case that you want to return false if you find an 'f', but you're returning true. Flip your true and false cases:

if (res.indexOf("f") != -1) {
   return false;
 } else {
   return true;
 }

ALSO your for loop is wrong because x starts at 0, so you need to go to < length not <= length of your string.

for (var x=0; x < arr[1].split("").length; x++) {

and now your code works as you wanted it to.

edited Nov 22, 2015 at 5:39

answered Nov 22, 2015 at 5:35

Leah Zorychta

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2 Comments

Samuel Cook Over a year ago

or better yet: return !res.indexOf("f")

splucena Over a year ago

@Macmee You're right I patterned my thinking on the first time I use indexOf and expected it to return true when I'm looking for false.

score 2 · Accepted Answer · 2015-11-22 05:49:42Z

2

Just edited your code. Click on the <p> to check:

function mutation(arr) {
  //return arr;
  res = "";
  for (var x=0; x< arr[1].split("").length; x++) {
    res += arr[0].indexOf(arr[1].split("")[x]) > -1 ? 't' : 'f';
  }
 return res.indexOf('f') > -1;
}

$('p').click(function(){
  alert(mutation(["hello", "hey"]));
  alert(mutation(["floor", "loo"]));
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<p>Click me</p>

edited Nov 22, 2015 at 5:49

answered Nov 22, 2015 at 5:44

user4090029

Comments

Community · Accepted Answer · 2017-05-23 11:52:20Z

0

If you simplify the logic a bit, that's easier to check:

function mutation(arr) {
  return arr[1].split('').reduce(function(res, x) {
    return arr[0].indexOf(x) >= 0;
  }, true);
}

Thanks Leon for the correction.

edited May 23, 2017 at 11:52

CommunityBot

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answered Nov 22, 2015 at 5:32

Hunan Rostomyan

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7 Comments

Leon Adler Over a year ago

does using true && condition have any benefit?

Hunan Rostomyan Over a year ago

@LeonAdler Compared to what?

splucena Over a year ago

This is working. Can you spot what's wrong with my code?

Leon Adler Over a year ago

compared to just condition - return arr[0].indexOf(x) >= 0

Hunan Rostomyan Over a year ago

@LeonAdler haha! Of course, thanks man. That was stupid of me.

|

Roberto Capelo · Accepted Answer · 2015-11-22 06:19:04Z

0

I tried to not chance your logic, the mistake are:

You're trying to compare with all characters on the array[0], not only the first.
If you find a character equals on the first character on array[0] you should return true.

Correct code:

function mutation(arr) {
  res = "";
  for (var x=0; x<=arr[1].split("").length; x++) {
    if (arr[0].split("")[0].indexOf(arr[1].split("")[x]) !== -1) {
      return true;
    }
  }
  return false;
}

answered Nov 22, 2015 at 6:19

Roberto Capelo

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