It is absolutely absurd some of the study plans that people go through where they are trying to study over multiple months. In order to truly memorize all the solutions, most people need these programs. Otherwise, it's just a luck of the draw. I actually got stuck at an interview because I forgot the nlogn solution for two sums. Absurd!
My favorite interview so far involved opening a raw TCP socket to Postgres and sending a query (Actually relevant to the job). I was given the prompt ahead of the interview and spent about 2 hours figuring it out. I learned something valuable and demonstrated an ability to expand my knowledge base. This interview has been the only one even remotely close to demonstrating my abilities to work at the job.
Are you talking about determining a pair of numbers in an array that sum to a given value? That's O(n) and just uses a hashset/hashmap.
This is one of those tricks you just have to memorize and it's very hard to come up with the solution in 30 min.
What a great way to kick off a professional relationship:
"I know this problem is useless and obviously unrepresentative of the actual work we do. So do you (if you aren't incompetent). You also know perfectly well that I've memorized the answer and am only pretending to 'solve' it for you on the spot. And yet, we go along with the charade and pretend it's a vitally necessary, even clever hiring technique. Because hey, we're getting paid big bucks to play this game, after all. So who cares."
Maybe that particular solution is hard to come up with, but you can solve the problem without any "tricks", just basic principles. I'll try to explain which principles I'd use using python.
You can start with the trivial O(N^2) solution:
def has_2sum(lst, target):
# returns whether there are 2 (not necessarily distinct) elements in `lst` which sum to target
for a in lst:
for b in lst:
if a + b == target: return True
return False
def has_2sum(lst, target):
for a in lst:
for b in lst:
if b == target - a: return True
return False
def has_2sum(lst, target):
for a in lst:
return (target - a) in lst
def has_2sum(lst, target):
set_lst = set(lst)
for a in lst:
return (target - a) in set_lst
import bisect
def has_2sum(lst, target):
sort(lst)
def contains(x):
# equivalent to `x in lst`
i = bisect.bisect_left(lst, x)
return (0 <= i < len(lst)) and (lst[i] == x)
for a in lst:
return contains(target - a)
Consider also that: 1-((1-.7)*3) = .97
3 interviews with a 70% chance give you a 97% chance of landing atleast one job. More interviews improve your situation rapidly. Job hunting is better seen as a campaign than individual battles.
But I also think maybe in my estimation you FAANG chances aren't that low as your 2%. They hire tons of people, ALL THE TIME.
What do you think your chances are if you're actually qualified? My made up gut numbers: At least 70% even if you don't practice leetcode. That's the real question here in my mind, what are an otherwise qualified candidates chances? How much does that change with interview skills, prep, leetcode etc.
Otherwise, it feels like a gamble of a huge time sink that could have gone towards something actually beneficial/profitable.
Having had too much of my time and energy drained in the past with the run around, I said fuck it the last time when facing the option to interview or go my own way, started my own thing and haven't looked back since.
Even when you do get an offer, it's a gamble on whether or not you're dealing with toxic management or not. Which only reinforces the idea of compensation for interviewing in case someone needs to jump that ship and start interviewing again.