Tip 1 : OOPS concepts are a must
Tip 2 : Graph and binary search are important concept because generally two medium DSA question will be asked
Tip 3 : Basic understanding of computer fundamentals is required
Tip 1 : Preferred work experience in backend or frontend
Tip 2 : Referral is required
There were three coding questions, All were of medium level, mainly including implementation, binary search, and DSU.



Given ‘N’ = 2, The total number of BST’s is 2.

1. A binary search tree is a rooted binary tree whose internal nodes each store a key greater than all the keys in the node's left subtree and less than those in its right subtree.
2. A structurally unique binary search tree is a tree that has at least 1 node at a different position or with a different value compared to another binary search tree.
There was a coding question, one was of binary search, and the other was a dfs implementation on graph



First I explained O(n) solution and then optimized it to Log(n).
Single DSA question, Parking lot system design, and computer fundamental questions from the operating systems, Computer networks, and OOPs concepts.


In this given binary tree, nodes 7, 5, and 9 are half nodes because they have only one child node. So we have to replace these nodes with their child. The inorder traversal of the updated binary tree is [1, 6, 11, 2, 4]. Hence, the answer is [1, 6, 11, 2, 4].

Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.
For example, the input for the tree depicted in the below image would be :
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level.
The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null (-1).
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
DFS

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
What is recursion?