☕ Message Queues
An awesome guide on Message Queues. Plus, what was it like to work under Steve Jobs in the 2000s as a Software Engineer at Apple? Also, an answer to yesterday's coding interview question.
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An awesome blog post on Message Queues
Covers things like
- What is a Message Queue and why are they necessary
- Delivery Mechanisms and the different delivery guarantees
- Ordering and Parallelism
- Popular message queues like AWS SNS & SQS, Google Pub/Sub, Redis Streams, etc.
An awesome podcast with David Shayer, a Software Engineer at Apple (he was there from 2001 to 2015).
There’s some interesting discussion on David’s work on file systems (HFS+) and his work with the iPod.
David was actually hired by Tony Fadell, who was in charge of the iPod and oversaw hardware and firmware for the iPhone.
Given two arrays of integers, compute the pair of values (one value in each array) with the smallest (non-negative) difference.
Return the difference.
Input - [1, 3, 15, 11, 2], [23, 127, 235, 18, 9]
Output - 2
Explanation - The difference between 11, 9
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As a refresher, here’s the last question
Given a positive integer n, write a function that computes the number of trailing zeros in n!
Input - 12
Output - 2
Explanation - 12! is 479001600
Input - 900
Output - 224
Explanation - 900! is a big ass number
This question requires a bit of basic number theory.
6! = 1 * 2 * 3 * 4 * 5 * 6 = 720
From this, we can see that the trailing 0 comes from 2 * 5 in our factorial equation.
We can generalize this to all factorials. Whenever we have a 2 * 5 in our factorial, then we will get a trailing zero.
This also applies to multiples 2 and 5. Whenever we have a multiple of 2 being multiplied by a multiple of 5, that will add a trailing zero. Or if we have a number that is a multiple of both 2 and 5.
Another way of thinking of this is breaking down each individual number in our factorial equation by its factors. Then we count the number of 2s and 5s and that will be the number of trailing zeros.
6! = 1 * 2 * 3 * (2 * 2) * 5 * (3 * 2)
We have four 2s but only one 5, so the number of trailing zeros is 1.
Multiples of 2 are more common than multiples of 5, so we can actually just count the number of 5s in our factorization break down and return that as the number of trailing 0s. We’ll always have enough 2s to match the 5s.
We can count the number of multiples of 5 by dividing n by 5.
However, this will undercount the number of 5s in our factorization when we have a number that has multiple 5s in it’s factorization.
For example, 25 = 5 * 5
So, we have two 5s in that factorization, but when we divide n by 5, we’re only counting 25 as a single 5.
Therefore, we have to repeat the process for all the powers of 5 and divide n by those powers so that we don’t undercount them.
Here’s the Python 3 code.
def trailingZeroes(n: int) -> int:
zeros = 0
power = 1
while 5**power <= n:
zeros += n // 5**power
power += 1
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