Position: Test Engineer [Automation]
Location: US (which later got changed to Bangalore)
Process took: 4 months

I applied through Google career portal in March and got mail from HR in a week. He asked me to schedule a call to start the interview process. Below is the detailed experience of every round.

HR phone screen:

In this call, the recruiter discussed the job description in detail and what is to be expected ahead. He asked me a few technical questions related to coding and testing (fyi, python is my main language).

1. Which data structure out of list, tuple and dictionary is immutable.
   Ans: Tuple
2. What is the best time complexity of quick sort?
   Ans: O(N log N)
3. What is the time complexity to read from of hash map?
   Ans: O(1)
4. How would you test Gmail application?
   Ans. Trick is to ask what kind of testing is to be performed and in what time limits. Whether it is just a feature addition or a completely new product. Think of security testing, performance, load (along with functional). Google more on this.

Towards the end, HR asked me to schedule my technical telephonic round in next 4-5 weeks as per my convenience.

Technical phone screen:

Write a function to return local time by taking city name as input. If input is California, then function should returns its local time in Hours:Minutes:Seconds format. Write test cases for it. How would you write automation test for this function.

from datetime import datetime
def get_gmt_time(city):
	now = datetime.utcnow()
	print(now)  #returns GMT time, like 2019-04-23 09:11:44.375078
	try:
		diff = gmt.getOffset(city)   #returns number of hours of offset for that city
	except:
		raise('Exception raised by API')
	dt = datetime.timedelta(diff, time=hours)
	return dt.hours,dt.minutes,dt.seconds,dt.milliseconds

Follow-up: how would I create automation test for this function. The solution was to make sure the test runs sharp at 9 AM IST (for eg.) and returns the expected answer everytime.

Onsite:

Round 1:

• How did I manage difficult situations at work
• When did I play leadership role at work and what were the challenges
• How did I deal with situation when somebody else gets credit for my work
• What would my manager say about my strengths and weaknesses
• How did I help a co-worker to solve problems
• How do I naviagate ambiguity at work when things are always changing

Round 2:

• https://leetcode.com/problems/shortest-path-in-binary-matrix

  My solution

  def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
  		n = len(grid)
  		if grid[0][0] == 1 or grid[n-1][n-1] == 1:
  			return -1
  		x_diff = [-1, -1, -1, 0, 0, 1, 1, 1]
  		y_diff = [-1, 0, 1, -1, 1, -1, 0, 1]
  		queue = collections.deque()
  		queue.append([0, 0, 1])
  		while queue:
  			i, j, length = queue.popleft()
  			if (i, j) == (n-1, n-1):
  				return length
  			l = ((i-1,j-1),(i-1,j),(i-1,j+1),(i,j-1),(i,j+1),(i+1,j-1),(i+1,j),(i+1,j+1))
  			for i, j in l:
  				if 0 <= i < n and 0 <= j < n and grid[i][j] == 0:
  					queue.append([i, j, length+1])
  					grid[i][j] = 1
  		return -1

Round 3:

• Decode string

  My solution

  I could not solve this problem in one pass as asked by interview. Nevertheless, posting the solution after searching online:

  from collections import deque
  def decodestring_easy(mystr):
  	i = 0
  	stack = []
  	while i < len(mystr):
  		c = mystr[i]
  		if c== '[':
  			stack.append(c)
  		elif c == ']':
  			temp = ""
  			while(stack[len(stack)-1]!='['):
  				temp = stack.pop() + temp
  			stack.pop()
  			stack.append(temp)
  		elif c == '{':
  			num = 0
  			i += 1
  			while i<len(mystr) and mystr[i] != '}':
  				num = num*10 + int(mystr[i])
  				i += 1
  			pre = stack.pop()
  			temp = ""
  			while num!=0:
  				temp = temp + pre
  				num -= 1
  			stack.append(temp)
  			# print(temp)
  		else: 
  			stack.append(c)
  		i += 1
  	res_str = ""
  	while len(stack)!=0:
  		res_str = stack.pop() + res_str
  
  	return res_str

Round 4:

• How to test a class StopWatch which measures elapsed time:

  class StopWatch {
  	private long startTime;
  	private long endTime;
  	public void start() {
  		startTime = System.nanoTime();
  	}
  	public void stop() {
  		endTime = System.nanoTime();
  	}
  	public long elapseTime() {
  		return endTime-startTime;
  	}
  }

• Write a function which takes a square matrix and zeroes out its both diagonals.

  My solution

  def operate(a):
  	n = len(a)
  	rows,cols = 0,0
  	while cols<=n-1:
  		a[rows][cols]=0
  		rows+=1
  		cols+=1
  	rows,cols=0,n-1
  	while cols>=0:
  		a[rows][cols]=0
  		cols-=1
  		rows+=1

Round 5:

• Write a function to find the 3rd largest number in a list of numbers https://leetcode.com/problems/kth-largest-element-in-an-array

  My solution

  My approach was to sort the list using merge sort and return third element.

  def pick_third_largest(A):
  n = len(A)
  if n<3:
  	return None
  mergeSort(A)
  return A[2]
  
  def mergeSort(A):
  	n = len(A)
  	if n>1:
  		mid = n//2 
  		L = A[:mid] 
  		R = A[mid:]
  		mergeSort(L)
  		mergeSort(R)
  		i=j=k=0
  		while i<len(L) and j<len(R):
  			if L[i] > R[j]:
  				A[k]=L[i]
  				i+=1
  				k+=1
  			else:
  				A[k]=R[j]
  				j+=1
  				k+=1
  
  		while i<len(L):
  			A[k]=L[i]
  			i+=1
  			k+=1
  		while j<len(R):
  			A[k]=R[j]
  			j+=1
  			k+=1

• Given a log file of following format, lines can be in any order or mixed up:

  Login 1: John Smith
  Login Success: 2
  Login 3: User123
  Login Failed: 3
  Login 2: Adam Apple
  Login Failed: 1

  Print out list of successful and failed logins:

  Success:
  Adam Apple
  Failed:
  User123
  John Smith
  My solution

  My approach was to parse each line in log file and maintain a hash table (dictionary) having user IDs and their corresponding names. Then, print out success and failures based on the names.

  dict_users=dict()
  success_list=[]
  failures_list=[]
  with open("log.txt") as f:
  	mylist = f.read().splitlines()
  	for line in mylist:
  		first,last = line.split(":")
  		userId_or_status= first.split(" ")[-1].strip()
  		if userId_or_status.isdigit(): 
  			dict_users[userId_or_status] = last.strip()
  		else:
  			if first.find("Success")!=-1:
  				success_list.append(last.strip())
  			if first.find("Failed")!=-1:
  				failures_list.append(last.strip())
  	print("Success:")
  	for user in success_list:
  		print(dict_users[user])
  	print("Failed:")
  	for user in failures_list:
  		print(dict_users[user])
  

Let me know if you have any alternative solutions for the problems. Would love to hear it.