Easy Uber Coding Interview Question - Check if a String Is an Acronym of Words - Leetcode 2828

FAANG Coding Interviews / Data Structures and Algorithms / Leetcode

[GregHogg]

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[tangocukedi]

Was the interview for a driver position?

[HiImKyle]

I just wanted a ride home, I didn't wanna have to answer any interview questions

[ceciljoel]

Is a candidate master rating level skill of codeforces enough to gurantee passing code interviews?

[Ridgwaycer]

Are we supposed to believe companies are giving questions which can be solved by baby's first for loop?

[brandondenis]

I was hoping to see someone suggest the .startswith() string method in python.

[rahuldwivedi]

Why do you need to check if a word at that index exists? Aren’t we already checking both lengths at the beginning itself?

[harshsinha]

From where are you picking up these leetcode questions?

[budimanmh]

How about if having an array of 4 words, but any of them could be empty, which could result the acronym have less chars compared to the number of words?

[hlubradio]

Always thumbs up. Just need some time to implement them. I did implement the perfect sq using binary search

[PhonesUnboxing]

I wish I get this question in an interview 😂

[vadymvyhovskyi]

programmer: 3!=4.
Mathematician: What?!

[ryugane]

how about:

try:
return s == "".join(word[0] for word in words)
except IndexError:
return False

Should still be O(n), I can understand relying on Exceptions might be criticizable but I feel like the added readability of it makes up for it

[movaxh]

To be pedantic. That does not check if it is acronym. It checks if it is an abbreviation. Also one liner in Python with zip, all and list comprehension.

[TradingBull]

So uber driver will ask this instead of otp

[jvcss]

I would like at least to pass to the technical interview. Can you share your advices for a stunning resume?

[Kaviarasu_S]

Submission works without the len(words[i]) == 0

[atomix]

Where can we find details for your 1 on 1 tutoring?

[rahuldwivedi]

That extra check len(words[i]) === 0 is actually unnecessary. Following would do fine.


var isAcronym = function(words, s) {
if(words.length !== s.length) return false;

for(var i=0; i<s.length; i++){
if(s[i] !== words[i][0]) return false;
}

return true;
};

[andrewbeaver]

Using the length to determine the validity of an acronyms, would be technically fail. There are acronyms that borrow the second of the word.

Example:
loran = LOng-RAnge Navigation.
SONAR
RADAR
WiFi ?!?! Literally one of the most common, two word list != string length of four