## Instructions

**shortest code: Bug in Apple
(Code length limit: 80 chars)**

This is the challenge version of coding 3min series. If you feel difficult, please complete the simple version

**Task**

Find out "B"(Bug) in a lot of "A"(Apple).

There will always be one bug in apple, not need to consider the situation that without bug or more than one bugs.

input: string Array apple

output: Location of "B", [x,y]

**Code length calculation**

In javascript, we can't get the user's real code, we can only get the system compiled code. Code length calculation is based the compiled code.

__For example:__

If you typed sc=x=>x+1

after compile, it will be:sc=function(x){return x+1;}

## My solution:

```
function sc(a){
for (i = 0; i < a.length; i++)
{
j = a[i].indexOf('B')
if (j > -1) return [i, j]
}
}
```

## Explanation

First I used a loop to iterate in the first array that contained the other arrays, and then inside it I declarated "j", that contained the index of 'B' inside the array that is being iterated, and after that I added a conditional that if the index of 'B' is higher than -1, it means that there is a 'B' element so I just returned the last result that is an array with the index of the array being iterated so I get the row and "j" that is the index of 'B', so I get the column

**What do you think about this solution? ðŸ‘‡ðŸ¤”**

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