Software engineer Interview Questions in Pittsburgh, PA

Talk about any challenges you faced and how you overcame them

Hello, Can I please reach you somehow. I need to ask some questions. I would be really thankful.

Can you please tell me about the kind of questions on your hackerrank interview and their difficulty and the technical questions in the onsite interview?

It's a standard bipartite matching problem, aka assignment problem. Solutions are standard as well, but none of them are easy. They cannot be figured out during the interview, you just have to know them. These types of questions just make me wonder about those who ask those questions. What exactly do they test? How long have you been from college where you studied Ford-Fulkerson maximum flow algorithm, or Hungarian algorithm (both applicable to the problem)? Show me an interviewer who can answer this question, unless he or she specially prepared it for an interview, thus wasting his productive time at work? And, by the way, trying to use textbook algorithms for real life assignment problems would lead to the results that are mediocre at best. Commercial-strength packages have lots of optimizations for special cases of input data. For instance, the Linux qsort command uses 3 different sorting algorithms depending on the size of input data.

maybe the interviewers just want to test whether you realize (in reasonable) time that this is a non-trivial problem to solve.

If one add can go into multiple slots then this is a simple problem. For every slot you can check which add gives maximum revenue. It is of order (m*n). Then how many adds can one slot take? If one slot can take as many adds as possible then you can put all the adds into most profitable slots leaving some empty. Again a simple solution. If an add can only go into one slot and each slot can take only one add (m>=n), then it is a complicated problem. May be interviewer wants to see if you can separate those cases.

Could you elaborate on the questions? Was it just OOP principles, or did they cover language specific topics?

An very easy question about overriding to compare objects. Do it on collabedit.

Did you use hacker rank as an online editor. Thanks

After indicating the naive approach, which is O(n^4), I produced a result that uses backtracking and a hash table to bring it down to O(n^3).

rectangles of 1s or 0s? or it doesnt matter? what does"and parallel to the array's dimensions" mean?

The solution would be the same either way. And "parallel to the array's dimensions" just basically means "don't worry about detecting diamonds or other perturbations".

def match(text, pattern): """ Given a string and a pattern, where: '.' - matches any single character. '*' - matches zero or more of the preceding element. Find the first substring matching this pattern. Tests: >>> match('alex', 'le') le >>> match('alexandra', 'al.*a') # an alexa >>> match('alexandru', 'al.*a.dru*') alexandru >>> match('alex', '.') 'a' >>> match('alex', '.*') 'alex' >>> match('aleex', 'e*') 'ee' >>> match('alex', 'alex*') 'alex' >>> match('alex', '*alex') Error >>> match('alex', 'alex.') '' >>> match('alex', '**') Error """ END = '$' class Automaton(object): """ Builds a new automaton for matching strings. """ def __init__(self, pattern): self.start = {} state = self.start symbols = self.extract_symbols(pattern) for (index, symbol) in enumerate(symbols): if len(symbol) == 1: state[symbol] = {} state = state[symbol] else: [x, y, z] = symbol state[x] = state if z != None: state[z] = {} state = state[z] state[END] = None def extract_symbols(self, pattern): """ Returns: list, of string, format ([A-z.]|([A-z.]\*[A-z]) """ out = [] i = 0 while i < len(pattern): c = pattern[i] n = pattern[i+1] if i+1 < len(pattern) else None nn = pattern[i+2] if i+2 < len(pattern) else None if c == '*' and i == 0: raise Exception('Pattern does not support * in first position') if c == '*' and n == '*': raise Exception('Pattern does not support successive *') if n == '*': out.append([c, n, nn]) i += 3 else: out.append(c) i += 1 return out def check(self, text, index): """ Checks if the string matches the automaton. """ text += END state = self.start while index < len(text): char = text[index] index += 1 if char in state: state = state[char] elif '.' in state: state = state['.'] else: return False return True a = Automaton(pattern) for i in range(len(text)): is_match = a.check(text, i) if is_match == True: return True return False

Could you please post clearer code? Or is it Glassdoor comments that are taking the whitespace away? I got the same two questions, but the second one was onsite. I cleared the interview at Facebook's Menlo Park office, but joined Google instead because they gave me a better team to join. (Aside: I got trained at InterviewKickstart.com. Check it out. Very legit. They are like a mini-CS school, with mock interviews)

When a higher priority task is preempted by a lower priority task, creating a scheduling issue.

What was the timeline involved in the hiring process? How long after your final interview did you hear from them?.

Implement a simple bit set

/* Java program to design a data structure that support folloiwng operations in Theta(n) time a) Insert b) Delete c) Search d) getRandom */ import java.util.*; // class to represent the required data structure class MyDS { ArrayList arr; // A resizable array // A hash where keys are array elements and vlaues are // indexes in arr[] HashMap hash; // Constructor (creates arr[] and hash) public MyDS() { arr = new ArrayList(); hash = new HashMap(); } // A Theta(1) function to add an element to MyDS // data structure void add(int x) { // If ekement is already present, then noting to do if (hash.get(x) != null) return; // Else put element at the end of arr[] int s = arr.size(); arr.add(x); // And put in hash also hash.put(x, s); } // A Theta(1) function to remove an element from MyDS // data structure void remove(int x) { // Check if element is present Integer index = hash.get(x); if (index == null) return; // If present, then remove element from hash hash.remove(x); // Swap element with last element so that remove from // arr[] can be done in O(1) time int size = arr.size(); Integer last = arr.get(size-1); Collections.swap(arr, index, size-1); // Remove last element (This is O(1)) arr.remove(size-1); // Update hash table for new index of last element hash.put(last, index); } // Returns a random element from MyDS int getRandom() { // Find a random index from 0 to size - 1 Random rand = new Random(); // Choose a different seed int index = rand.nextInt(arr.size()); // Return element at randomly picked index return arr.get(index); } // Returns index of element if element is present, otherwise null Integer search(int x) { return hash.get(x); } } // Driver class class Main { public static void main (String[] args) { MyDS ds = new MyDS(); ds.add(10); ds.add(20); ds.add(30); ds.add(40); System.out.println(ds.search(30)); ds.remove(20); ds.add(50); System.out.println(ds.search(50)); System.out.println(ds.getRandom()); } }