FAANG Level Mock Software Engineer Interview (JavaScript)

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I hosted a 3rd and final mock interview with this series of FAANG-level coding challenges. Daniel, the interviewer, has almost 20 years of experience as an engineer between Microsoft and Facebook. We found a brave soul to be the interviewee and he actually got to the final challenge in this interview. If you want to see what the coding challenge of a FAANG level software engineer interview looks like, this episode is for you. Enjoy!

Sections:
0:00 - Intro
3:57 - 1st coding challenge
18:28 - 2nd coding challenge
48:19 - Questions and feedback
59:23 - Outro

Daniel Tomko (guest):

Brian Do (guest):

[varuntyagi]

That was a really nice interview. Shout out to Daniel for keeping it friendly and lighthearted the whole time. I've been in interviews where the interviewer was very rigid and it in-turn made me very nervous.

[ozzyfromspace]

This was actually a really good interview! Well done Brian!

[pmulard]

Just want to give a shout out to Daniel, who kept the interview lighthearted and fun :) I'm sure it goes a long way in easing the interviewee's anxiety and bringing out the best in them.

[jscul]

It's funny this interview. At 10:25, the interviewer is actually wrong, you don't need quotes in JS. You can just do {a: 1, b: 2}. Python is the version where you'd need quotes. I'd probably comment on that and explain that in JS if you want to name something a key using a variable you would do {[a]: 1, [b]: 2}.

[omgalul]

Thanks for the awesome interview!

I was thinking, create trie out of words --> create freq map --> evaluate every word in the trie while using backtracking to undo modifications to the freq map.

The efficiency here is that you don't recreate the freq map for each word + stopping early and not repeating work when evaluating words with same prefixes like 'dog' and 'dogs' etc..

run time: O(2D + T ---> D + T) D is number of letters in trie, T number of tiles in freq map

[austinlawan]

Thank you for these sessions guys. They very very helpful.

[laszloekovacs]

I'm way too bad as a programmer but I managed to solve this with map, reduce. One of my proudest moments.

[lailbeeb]

Daniel was awesome. It's better to have lighthearted goofy interviewer than stone faced interviewer who says nothing, I hate those interviews.

[SpinozicTroll]

Wow what an amazing interview, thank you so much for your content.

[pedroreyes]

Great interview and video!

I felt like mentioning the optimization for the game of scrabble (wordbook wont change, can be fully processed) is kind of out of the scope of the problem at hand.

[mrawesome]

Hey Don,

Would you be willing to do interviews with some people are aspiring developers? I think it would be beneficial to interview those who are working to becoming a developer.

LOVE THE VIDEOS!

[trignal]

It's impossible to read the question. Can someone please post it here?

[alexjosesilvati]

Very good interview ingles!
Is helping the interview...

[qaqiaq]

Feel like the first answer was over engineered/hard to read. Here is what I came up with:

const scoreWord = (word, tiles) => {
let result = 0

for (const char of word) {
if (tiles.includes(char)) {
result += alphabet[char]
tiles = tiles.replace(char, '')
} else if (tiles.includes('_')) {
tiles = tiles.replace('_', '')
} else {
return 0
}
}

return result
}

and for second question:



const scoreWords = (words, tiles) => {
let score = 0
let bestWord = ''
for (const word of words) {
let tempTiles = tiles
let tempScore = 0
if (word.length > tiles.length) {
return
}

for (const char of word) {
if (tempTiles.includes(char)) {
tempScore += alphabet[char]
tempTiles = tempTiles.replace(char, '')
} else if (tempTiles.includes('_')) {
tempTiles = tempTiles.replace('_', '')
} else {
continue
}
if (tempScore > score) {
score = tempScore
bestWord = word
}

}
}
return [score, bestWord.length > 0 ? bestWord : null]
}

console.log(scoreWords(['cat', 'dog', 'foo'], 'cado_f')) //[5, "foo"]

[allensun]

And leetcode reference of the second problem?

[jjescandor]

Here's my solution to the second problem:
def max_score(list_str, tiles):
max = 0
word = ""
for str in list_str:
total = scorer(str, tiles)
if total > max:
max = total
word = str
return [max, word]

print(max_score(["cat", "dog", "dogs", ], "tmac"))

[JJ-psvb]

Isn't the time complexity of Q2 the same as checking the score for every word? He's just pruning the search but not improving the time complexity.

[kevinnguyen]

Great practice interview and great job to Brian! Was a interesting problem, as I also never played Scrabble. Was also fun to try it myself and ended up getting a different solution.

Thanks for sharing!

const dictPoints = {
a: 1,
b: 3,
c: 3,
d: 2,
e: 1,
f: 4,
g: 2,
h: 4,
i: 1,
j: 8,
k: 5,
l: 1,
m: 3,
n: 1,
o: 1,
p: 3,
q: 10,
r: 1,
s: 1,
t: 1,
u: 1,
v: 4,
w: 4,
x: 8,
y: 4,
z: 10,
_: 0,
};

function getDictionary(object, key) {
let result = object[key];
return result;
}

function checkMap(tiles) {
let map = {};

for (let i = 0; i < tiles.length; i++) {
if (!getDictionary(map, tiles[i])) {
map[tiles[i]] = 1;
} else {
map[tiles[i]] += 1;
}
}

return map;
}

function calculateWord2(array, tiles) {
sum = 0;
bestPoints = {};
map = checkMap(tiles);

array.forEach((string) => {
sum = 0;
for (let i = 0; i < string.length; i++) {
if (getDictionary(map, string[i])) {
sum += getDictionary(dictPoints, string[i]);
map[string[i]] -= getDictionary(map, string[i]) - 1;
}
bestPoints[string] = sum;
}
});

return bestPoints;
}

console.log(calculateWord2(["cat", "dog", "test"], "tke_skdac"));

[ericnail]

The only thing that bothered me was that he was correct about copying the letters, all object keys are inherently strings and those keys were all valid var names in JavaScript. No quotes needed.

Really cool interviewer, though. A+

[mshawn]

I didn't understood the first question only can anyone explain it to me