Questions d'entretiens - Back end engineer

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Veeva Systems
On a demandé à Senior Software Engineer, Java Back End...18 août 2015

public class Person { Person father; Person mother; Gender gender; Integer age; List<Person> children; int level = 0; public enum Gender { Male, Female; } } For the above class, you basically have to implement 2 methods. public List<Person> getOldestSisters() public List<Person> getGreatestAncestors()

5 réponses

/** * Returns the Persons which are the greatest number of levels up the family tree * from this instance. Examples: * Given a tree where this person instance only has a mother and father, return mother and father. * Given a tree where this person instance has a mother &amp; father, and a grandfather on the mother's side, return only the grandfather on the mother's side. * @return List of Person */ public List getGreatestAncestors() { // If this is the root of the tree, return empty array because there is no ancestor for the tree if (this.father == null &amp;&amp; this.mother == null) { return new ArrayList(); } List fList = new ArrayList(); List mList = new ArrayList(); if (this.father != null) { fList = this.father.getGreatestAncestors(); } if (this.mother != null) { mList = this.mother.getGreatestAncestors(); } List results = new ArrayList(); for (Person p : fList) { if (results.contains(p)){ continue; } results.add(p); } for (Person p : mList) { if (results.contains(p)){ continue; } results.add(p); } return results; } Moins

I cranked this out in about 30 minutes. I believe it works quite well. I've also included the corresponding unit tests: file: Person.java ------------------------------------------------------------------------------------------------------------- package Command; import java.util.ArrayList; import java.util.HashSet; import java.util.List; public class Person { Person father; Person mother; Gender gender; Integer age; List children; int level = 0; public enum Gender { Male, Female; } Person(Person dad, Person mom, Gender gender, int age, int level) { this.father = dad; this.mother = mom; this.gender = gender; this.age = age; this.level = level; } public void setChildren(List children) { this.children = children; } public void addChild(Person child) { this.children.add(child); } public List getOldestSisters () { // given the current person (self), determine parents // then get children of those parents // Determine gender of each child // Where they are female, get ages // return females with age &gt; mine // Note: must check on both sides father/mother as there may be a mixed marriage // Combine list of children - Exclude YOU as you cannot be your own sister. // Use a set to eliminate duplicates. HashSet allChildren = new HashSet(); // Can't add null to a hashSet so screen for it. if ((father != null) &amp;&amp; (father.children != null)){ allChildren.addAll(father.children); } if ((mother != null) &amp;&amp; (mother.children != null)) { allChildren.addAll(mother.children); } // If you are not in this list, there is an issue! if (allChildren.contains(this)) { allChildren.remove(this); // System.out.println("Removing self from list."); } else { System.out.println("Error: You are not a child of your parents! Adopted?"); } // Filter down to only women and get any older than me: int myAge = this.age; List oldestSisters = new ArrayList(); for (Person child : allChildren) { if (child.gender == Gender.Female) { if (child.age &gt; myAge) { oldestSisters.add(child); } } } return oldestSisters; } public List getGreatestAncestors() { if ((this.father == null) || (this.mother == null)) { return null; // You must have two parents to have ancestors } // Find root parents List myParents = getParents(this); return getElders(myParents); } private List getElders(List parents) { List elders = new ArrayList(); List myParents = new ArrayList(); boolean newElders = false; for (Person parent : parents) { myParents = getParents(parent); if (myParents.isEmpty()) { elders.add(parent); } else { elders.addAll(myParents); newElders = true; } } if (newElders == true) { return getElders(elders); } return elders; } private List getParents(Person person) { List parents = new ArrayList(); if (person.father != null) parents.add(person.father); if (person.mother != null) parents.add(person.mother); return parents; } } // For the above class, you basically have to implement 2 methods. // public List getOldestSisters() // public List getGreatestAncestors() Moins

package Command; import org.junit.jupiter.api.AfterEach; import org.junit.jupiter.api.BeforeEach; import org.junit.jupiter.api.Test; import java.util.ArrayList; import java.util.List; import static org.junit.jupiter.api.Assertions.*; class PersonTest { // Create grand parents Person grandFatherDad = new Person(null, null, Person.Gender.Male, 78, 1); Person grandMotherDad = new Person(null, null, Person.Gender.Female, 81, 1); Person grandFatherMom = new Person(null, null, Person.Gender.Male, 69,1); Person grandMotherMom = new Person(null, null, Person.Gender.Female, 89, 1); // Create parents / aunts and uncles Person father = new Person(grandFatherDad, grandMotherDad, Person.Gender.Male, 48, 2); Person mother = new Person(grandFatherMom, grandMotherMom, Person.Gender.Female, 47, 2); Person aunt1 = new Person(grandFatherMom, grandMotherMom, Person.Gender.Female, 47, 2); // Twin sis to mom Person aunt2 = new Person(grandFatherDad, grandMotherDad, Person.Gender.Female, 37, 2); Person uncle1= new Person(grandFatherDad, grandMotherDad, Person.Gender.Male, 35, 2); // Children (me and bros and sis) Person self = new Person(father, mother, Person.Gender.Male, 18, 3); Person brother1 = new Person(father, mother, Person.Gender.Male, 16, 3); Person sister1 = new Person(father, mother, Person.Gender.Female, 15, 3); Person sister2 = new Person(father, mother, Person.Gender.Female, 14, 3); @BeforeEach void setUp() { // Create children / sibling groups List children = new ArrayList(); children.add(father); children.add(aunt1); children.add(uncle1); grandFatherDad.setChildren(children); grandMotherDad.setChildren(children); children.clear(); children.add(mother); children.add(aunt2); grandFatherMom.setChildren(children); grandMotherMom.setChildren(children); children.clear(); children.add(self); children.add(brother1); // Dad had brother with other wife children.add(sister2); father.setChildren(children); // mom came with her own daughter from prior marriage children.clear(); children.add(self); children.add(sister1); children.add(sister2); mother.setChildren(children); } @AfterEach void tearDown() { } @Test void getOldestSisters() { // When there are no older sisters and I am male: //Person me = new Person(father, mother, Person.Gender.Male, 48); List olderSisters = null; olderSisters = self.getOldestSisters(); assertTrue(olderSisters.isEmpty(), "No older sisters."); // When there is one older sister and I am male Person sister3 = new Person(father, mother, Person.Gender.Female, 50, 3); mother.addChild(sister3); father.addChild(sister3); olderSisters = self.getOldestSisters(); assertTrue(olderSisters.contains(sister3)); assertTrue(olderSisters.size() == 1, "One older full sister."); // Youngest Sister has two older sisters (one full, one half) olderSisters.clear(); olderSisters = sister2.getOldestSisters(); assertTrue(olderSisters.contains(sister1)); assertTrue(olderSisters.contains(sister3)); assertTrue(olderSisters.size() == 2, "One older full, one older half"); } @Test void getGreatestAncestors() { List ancestors = self.getGreatestAncestors(); assertTrue(ancestors.size() == 4); assert(ancestors.contains(grandFatherDad)); assert(ancestors.contains(grandFatherMom)); assert(ancestors.contains(grandMotherDad)); assert(ancestors.contains(grandMotherMom)); ancestors = father.getGreatestAncestors(); assertTrue(ancestors.size() == 2); assert(ancestors.contains(grandFatherDad)); assert(ancestors.contains(grandMotherDad)); ancestors = mother.getGreatestAncestors(); assertTrue(ancestors.size() == 2); assert(ancestors.contains(grandFatherMom)); assert(ancestors.contains(grandMotherMom)); ancestors = grandFatherDad.getGreatestAncestors(); assertNull(ancestors, "getGreatestAncestors():Persons with no ancestors should return null."); } } Moins

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Zalando

round 1: Why coming to Berlin from India? Why Zalando? What do you know about Zalando? Salary expectations? Do you have any questions? round 2: Tell me about about a situation when there has been a conflict between you and your colleague, how you resolved it. Tell me about about a situation when there has been a conflict between two of your colleagues, they approached how you resolved it. How do you convince people to choose a technology? Tell me about about a situation when you went against odds / didn't listen to your manager. How you convince people to help you with technical issues. Tell me about about a situation which proves you are motivated to take up challenges and learn new technologies. How you choose a a particular technology/solution from among multiple technologies/solutions.

4 réponses

They will be testing your interpersonal and motivational skills.

round 3 questions: reverse a string recursively. explain architecture of Cassandra (I worked on Cassandra in previous projects) explain consistency configuration in Cassandra. Moins

Could you please post the questions that were asked to you and the kind of concepts that were touched upon, as part of the last 3 hour interview round? Moins

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Punch (CA)

They wanted a 2 minute profile video, for their clients

3 réponses

This isnt online dating!

The same thing happened to me. Is this a scam?

I get a weird vibe from this company. Not even sure they exist for real.

Rakuten

The Codility Test was about taking an integer and determine how many different arrangements of the number are possible. So if given, 132, arrangements are 132, 123, 231, 213, 312, 321. So the function should return 6.

3 réponses

Python solution: import itertools def solution(N): s = list(str(N)) solution = set([ s for s in itertools.permutations(s)]) if len(solution) == 1: print("1") return final_solution = [ s for s in solution if s[0] !='0' ] print(len(final_solution)) Moins

import collections def fact(n): factorial = n for i in range(1, n): factorial *= i return factorial def better_solution(N): s = list(str(N)) string_counts = collections.Counter(s) print(string_counts) permutations = fact(len(s)) # dealing with duplicate string values for string, count in string_counts.items(): if count &gt; 1: permutations //= fact(count) return permutations Moins

#include #include using namespace std; int factorial(int n){ return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n; } // Returns number of occurences of char c in string s int numberOfOccurences(string s, char c){ int occurences = count(s.begin(), s.end(), c); return occurences; } int solution (int N){ string numString = to_string(N); int numOfDigits = numString.length(); int computation = factorial(numOfDigits); // Using ASCII Codes for (int i = 48; i &lt; 58; i++) { char currentChar = (char) i; int occurences = numberOfOccurences(numString,currentChar); if (occurences != 0){ computation = computation / (factorial(occurences)); } } return computation; } Moins

Nordeus

How to determine which button offers which drink from a vending machine, knowing that all the labels are mixed.

2 réponses

The first thing that came to mind was Bubble Sort, but other (and more efficient) sorting methods should work too. Moins

First the questioned was not defined very well by the interviewer, second the way how the problem was declared did not cover all the possible cases! Moins

Yelp

Some questions are not that clear . Eg. How can you find all instances of a class. I did not understand that.

2 réponses

OR on runtime: {v for v in locals().values() if isinstance(v, SomeClass)}

the question was, how can you find all instances of a class in a file. Answer would be "grep". Moins

Grow Technologies

You are given a 5L and a 3L jug. Both jugs are really oddly shaped and don't have any measure markings, so you can only fill the jug up to the top. Given unlimited water, can you somehow make 4L of water from these two jugs?

2 réponses

This was the Die Hard III problem!

This is a classic question that is in CTCI/CareerCup. There are two solutions for this. 1) Fill up the 5L jug, then pour it into the 3L jug (5L jug has 2L of water in it). Pour out the 3L jug and pour the 5L jug into the 3L jug (3L jug now has 2L in it, 5L jug has 0L). Fill up the 5L jug with water, and pour it into the 3L jug until it is full! (5L jug has 4L in it, 3L jug has 3L in it) 2) Fill up the 3L jug, then pour it into the 5L jug (5L jug has 3L in it, 3L jug has nothing). Fill up the 3L jug again, and pour it into the 5L jug until it is full (5L jug has 5L in it, 3L jug has 1L in it). Pour out the 5L jug, then pour the 3L jug into the 5L jug (5L jug has 1L in it, 3L jug has 0L). Fill the 3L jug and then pour it into the 5L jug (5L jug has 4L in it, 3L jug has 0L). Moins

News360

Can you talk about one of your developing experience?

2 réponses

I have used python to develop a website application

During my last year of bachelor's of engineering I was working on mobile application project. I have used java and sql database to create and adroid application called Person Chat Mobile Moins

RetailMeNot

SQL: you have a table with customers, with gender m and f. Write a query to update m with f, f with m in a single query, without temporary tables

2 réponses

UPDATE customers SET gender = CASE WHEN gender = 'm' THEN 'f' WHEN gender = 'f' THEN 'm' END Moins

UPDATE customers SET gender = CASE WHEN gender = 'm' THEN 'f' WHEN gender = 'f' THEN 'm' END Moins

Fanatics

"Suppose you have two different kinds of server requests - requests that MUST be responded to in 2 seconds and cannot be resent, and requests that can be responded to in 10 seconds, but can be resent. How would you handle the priority of responding to these requests?"

2 réponses

Take prior constantly

I got it wrong a couple times, but with guidance from the interviewer, eventually came to the conclusion that priority for the short-time requests should start out higher than long-time requests, but that a long-time request nearing its timeout window should rise in priority to the point that it can be handled before a new (and unlikely to time-out) short-time request. Moins

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