J'ai postulé via la recommandation d'un employé. Le processus a pris 1 semaine. J'ai passé un entretien chez Whisper (Venice, CA) en juin 2017
Entretien
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.
J'ai postulé via un recruteur. J'ai passé un entretien chez Whisper (Los Angeles, CA) en avr. 2017
Entretien
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.
Questions d'entretien [1]
Question 1
Multiple string manipulation examples and problems.
J'ai postulé en ligne. Le processus a pris 2 jours. J'ai passé un entretien chez Whisper en janv. 2016
Entretien
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.
Questions d'entretien [1]
Question 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
