Senior SWE at Google, Amazon

Thank you for the interview. The time we spent together was very helpful! My interviewer joined and asked if we could reschedule our prep interview to an hour later. Since I really needed the prep interview I was flexible to the prep-interviewer, but it was inconvenient for me because I had to cancel other plans.

Thank you *******! Your 5-min silence tip is extremely helpful and I will practice graph+DP problems as you suggested. I definitely prefer your solution to the problem because it is more structured. However, my solution actually works, although it might not be the most suitable solution for the interview. Here is the working code that I finished within 10 mins after the interview. class Node: def __init__(self, value): self.left = None self.right = None self.parent = None self.value = value def inorder_recur(root, result): if root: inorder_recur(root.left, result) result.append(root.value) inorder_recur(root.right, result) def inorder(root, result): # start with the left most child of the root node_here = root while node_here.left: node_here = node_here.left result.append(node_here.value) # go to next node while node_here: node_here = go_to_next_node(node_here) if node_here: result.append(node_here.value) def go_to_next_node(node_here): # check if has right child, if so, return the left-most child of the right-child, or the right child itself if node_here.right: node_here = node_here.right while node_here.left: node_here = node_here.left return node_here # otherwise return the 1st right parent else: while node_here.parent and node_here.parent.left != node_here: node_here = node_here.parent if node_here.parent and node_here.parent.left == node_here: return node_here.parent else: return None root = Node(0) n1 = Node(1) n2 = Node(2) n3 = Node(3) n4 = Node(4) n5 = Node(5) n6 = Node(6) n7 = Node(7) n8 = Node(8) root.left = n1 n1.parent = root root.right = n2 n2.parent = root n1.right = n3 n3.parent = n1 n2.left = n4 n4.parent = n2 n3.left = n5 n5.parent = n3 n3.right = n6 n6.parent = n3 n6.left = n7 n7.parent = n6 n6.right = n8 n8.parent = n6 result1 = [] inorder(root, result1) print(result1) result2 = [] inorder_recur(root, result2) print(result2) assert result1 == result2 The expert provided very on-point tips. Very clear communication.

Interviewer gave me a good deep dive about Redis cache which I very much appreciated. I think for some reason the interviewer thought I was interviewing for Google. I know this is a senior level and also a mock, but it might be helpful if interviewer wanted me to deep dive into something, he asked. System design is very open ended and lot to cover. Little bit more direction would have been helpful but otherwise appreciated the session

He knows things really well, and enjoyed the discussion. Given that I am taking the services again, will like to interview with him :). I learned about certain aspects of sql vs nosql and how to go about the design. Thanks, ******* for your time :). Wish you all the best!

Nice experience!

Got what was advertised.

Excellent interviewer

Very useful, but a bit expensive - perhaps discount for users booking multiple sessions?

Thanks for the interview! Process went well. No issues.

Thank you for your feedback! Prepfully service is useful and the feedback received is always helpful

Very good interview quality.

It equipped me to behave professionally and prepare more for the coming interviews.

Great session. Didn't realize we'd be using google docs for writing code.

It was useful and I think most people can benefit from it.