Notes for a mock interview¶
- self intro infrastructure engineer in search ml infrastructure, inside tools for testing and developments service owner for search slogan service team search feature - for users and other engineering teams drive the team of search quality - responsible organize and manage the developments and teams - details
impact, complexity for what you have done technically!! millions level users and high qps, autoscaling and analysis tech stacks - details
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Why do you want to change your company? food truck search problem 没听清楚 → double check
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What is the biggest bug/mistake you have committed in your work? broken for the owned service details - main service -> microservice, parameters - coordination of the dependency of push process for different time zone - Why - deep analysis What -
3.1 how do you solve the problem - oncall wake up 2AM - push and rollback in time 3.2 what do you learnt and what to do in the future - use the centralized timezone - how do you enforce? What are the actions? - change the whole process, share calendar, ping on slack and calendar, link on other events, mark and account for the completeness - work and cooperate with diverse working environment - proactively communication
Literal point 0
insight point 1 - why your interviewer asked this question? Passion to join company?
insight point 2 - think about in depth what he wants to know? What can you bring to our company?
转专业的问为啥转码应该怎么回答 ??
insight point 1 - passion + motivation ⇒
100 / 3
100 -3-3-3-3-3 …..
10^10 / 1
Given two int value a and b, find a/b Notice: you can not use "/" operation.
4 / 3 = 1 --- 1
4 / 3 = 1
x * 3 = closet number to 4 but smaller than 4
< 0 sign 1 or both > 0 sign 1, -1 Math.abs(a) Math.abs(b)
if (a == 0) { return 0; }
min = 1, max = dividend mid = (max - min) / 2 + min mid * divisor = number > dividend max = mid; mid * divisor = number < dividend min = mid + 1 until we make min == max min will be the closest we can and smaller than dividend
dividend = 4, divsior = 3
min = 1, max = 4
mid = (4 - 1) >> 1 + min = 2;
number = 2 * 3 = 6 > 4 max = mid = 2
min = 1, max = 2 mid = 1 number = 1 * 3 = 3 < 4 min = mid + 1 = 2; min = 2, max = 2
we get result as 2 we check 2, 2 * 3 = 6 we will return 2 - 1 = 1
min = mid max = mid - 1
min =1 , max = 4 mid = 2 max = 2 - 1 min = 1, max = 1
public int divide(int dividend, int divisor) { if (divisior == 0) { throw new Exception(“divisor cannot be 0”); } if (dividend == 0 || dividend < divisor) { return 0; } int min = 1, max = dividend; while (min < max) { // when min equals max, end it int mid = (max - min) >> 1 + min; // get mid value int number = mid * divisor; // calculate possible value if (number == dividend) { // we find it, so we directly return it return mid; } else if (number > dividend) { // it looks max = mid - 1; } else { min = mid; } } return min; }
dividend = 10, divisor = 5
min = 1, max = 10
mid = 5 5 * 5 = 25 > 10
min = 1, max = 4 mid = 2 5 * 2 = 10 = 10 return 2
dividend = 4, divisor = 3
dividend = 0 return 0
divisor = 0 throw out of exception
1 / 2 dividend < divisor 0
dividend = 11 divisor = 3
min: 1 max: 11 mid = 6 * 3 = 18 > 11
min: 1 max: 5 mid = 3 * 3 = 9 < 11 min: 3 max: 5 mid = 4 * 3 = 12 > 11 min: 3 max: 3 mid = 3 * 3 = 9 < 11 mi
1 -- dividend log(dividend) O(1)
in the worst case, 1 2 ^ 32 ---> 2^31 --->
10^10 / 10^9 = 10 1 0 1 0
search range - [1, 2^4]
min: 1 to max: 10^10
log(a/b) = loga - logb
Clarification & Assumptions: b = 0? Why throw an exception in this case? - this is runtime error NullPointerException ArrayIndexOutOfBoundException IlegalArumentException IlegalStateException ArithmeticException ("divisor can not be 0")
types of a,b - int, overflow? the result will never overflow?
MIN_VALUE = -2^31
MAX_VALUE = 2^31 - 1
MIN_VALUE / -1 = ?
What is the result type? int
positive/negative for a,b?
-7 / 4 = -1 or -2
Result:
Binary Search Application
动手过例子，--->
1. 逻辑的关键部分 整理清楚 各种case，物理意义，
2. 解释你的high level，detail的逻辑
4 / 3
1 2 3 4
3 6 9 12
min mid max
2*3 =6
min,max
high level - 为什么能用binary search
把Binary Search的要点讲清楚 1. search range → [min, max] 2. search range reduce size each round → while loop()能进得去和跳的出来
[1, 2]
min max
mid
case 1 max = mid - 1 case 2 min = mid
Test cover all possible code branches
10 / 3 = 3