[Ber-FT]Finished Sebastian - Martin #629
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| hacker2Backwards += hacker2[i - 1]; | ||
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| console.log(hacker2Backwards); | ||
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All good in the hood!
A small comment:
For the reverse loop, instead of doing hacker2[i-1] you could change it in for (let i = hacker2.length - 1; i >= 0; i--);
which in terms of common practice in js would be that way, but it still works and there is no problem with your code.
Another small note: When naming variables for loops such as "let i" remember you can use other names for the variables in the loops such as "let j".
| console.log("number of et: " + counter); | ||
| console.log("number of words: " + wordCounter); | ||
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Nice job guys!
I guess you guys were counting "of" instead of "et" but looks like you got the idea of the lab.
A small note on the "of" counter: If there was a word like "software" your counter will count it as an extra "of" when actually there isn´t. So a small correction would be to make sure there is a space after the "f" in your code
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That's actually what we went for, we decided that we should count every "of" regardless if it is found in word or by itself.
I guess we could do if (words[i] == "o" && words[i + 1] == "f" && words[i + 1] == " ") . right?
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Or should it actually be : if (words[i] == "o" && words[i + 1] == "f" && words[i + 2] == " ") . ?
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