Ho presentato la mia candidatura tramite segnalazione di un dipendente. La procedura ha richiesto una settimana. Ho sostenuto un colloquio presso Whisper (Venice, CA) nel mese di giu 2017
Colloquio
I had a phone screen with someone, and then I flew out the next week for several in person interviews. I also had lunch at the office with several employees and then heard back from them the next day.
Ho presentato la mia candidatura tramite un selezionatore. Ho sostenuto un colloquio presso Whisper (Los Angeles, CA) nel mese di apr 2017
Colloquio
3 step interview. Started with a basic screening, asked about experience and engineering background. Then specialized questions and a coding example using some online toool (cant remember which one). The last one was more HR focused. Pretty straightforward process and very transparent.
Domande di colloquio [1]
Domanda 1
Multiple string manipulation examples and problems.
Ho presentato la mia candidatura online. La procedura ha richiesto 2 giorni. Ho sostenuto un colloquio presso Whisper nel mese di gen 2016
Colloquio
I applied online and a recruiter reached out to me to schedule phone screen with an engineer. I didn't continue in the process because I didn't pass the inital phone screen.
Domande di colloquio [1]
Domanda 1
1. Calculate how many times 10 can by divisible by a number?
- Gave an O(n) Solution:
public int count(int n){
if(n == 0){
return 1;
}
int count = 0;
while(n % 10 == 0){
count++;
n /= 10;
}
return count;
}
2. How can you do better than linear?
- Was a lot more difficult for me, but finally got an O(log n) Solution
public int binarySearch(int val, int start, int end){
if(start > end){
return -1;
}
int mid = (start + end) / 2;
if(val % 10^(length - i) == 0 && ((val / 10^(length - i)) % 10) != 0){
return (length - i);
}
else{
// if value is non 0, search right for value
// if value is 0, search left for value
if(val % 10^(length - i) != 0){
binarySearch(val, mid+1, end);
else{
binarySearch(val,start, mid-1);
}
}
}
3. Given 2 people, find if one is the ancestor of another? These are actual people so, there is a mother and father who are ancestors of their son, along with grandmothers and grandfathers. (Not a tree)
Given:
def isAncestor(me, anotherPerson):
class Person:
string id
string name
int gender
date dob
- What if the path up the tree is very long? It takes a long time to traverse up?
- Can store additional info in the Person class
