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'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.

• javascript

• what exactly you want to do? – Mritunjay Commented Nov 22, 2015 at 5:27
• I just pasted your code into the JS console and I got true as well – csga5000 Commented Nov 22, 2015 at 5:28
• If you are using IE8 and olders, .indexOf() doesn't be supported. Reference link: stackoverflow./questions/3629183/… – user4090029 Commented Nov 22, 2015 at 5:32
• im using chrome and firefox. – splucena 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 Commented Nov 22, 2015 at 5:41

4 Answers 4

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.

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./ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<p>Click me</p>

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.

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

• You're trying to pare 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;
}