Senior Software Engineer Interview Questions

List 1: tree nodes inorder List 2: tree nodes postorder List 3: all the nodes in between the given 2 nodes List 4: all the nodes after the given 2 nodes the LCA is the common node in List 3 and List 4.

I think there's an easier solution: List1: the parents of node1 in order bottom to top (can get them by navigating up tree from node 1). Navigate up tree from node2. First parent of node 2 found in list1 is LCA.

Use MergeSort.

You don't need merge sort here. This is a O(n) task, because the lists are already sorted. In fact, this merge routine is part of merge sort. No sorting takes place here, only merging, and it could be done in linear time. Node* merge(Node* list1, Node* list2) { Node* merged = null; Node** tail = &merged; while (list1 && list2) { if (list1->data data) { *tail = list1; list1 = list1->next; } else { *tail = list2; list2 = list2->next; } tail = &((*tail)->next); } *tail = list1 ? list1 : list2; return merged; }

Put me in a very unconformable situation.

Do you realize I'm applying for a software engineer position?

simple algorithm

Do you have any experience with KPIs?

Everyone who knows what an of statement is has written fizz-buzz. Check x%3, then check x%5, and write out your four if/else cases. You can use only a single branch with a switch on from x%15 from -14 to 14, or no branches at all by hardcoding 30 strings in an array and accessing that index.