SDE - 1

The city of Ninjaland is analogous to the unweighted graph. The city has ‘N’ houses numbered from 1 to ‘N’ respectively and are connected by M bidirectional roads. If a road is connecting two houses ‘X’ and ‘Y’ which means you can go from ‘X’ to ‘Y’ or ‘Y’ to ‘X’. It is guaranteed that you can reach any house from any other house via some combination of roads. Two houses are directly connected by at max one road.

A path between house ‘S’ to house ‘T’ is defined as a sequence of vertices from ‘S’ to ‘T’. Where starting house is ‘S’ and the ending house is ‘T’ and there is a road connecting two consecutive houses. Basically, the path looks like this: (S , h1 , h2 , h3 , ... T). you have to find the shortest path from ‘S’ to ‘T’.

For example

In the below map of Ninjaland let say you want to go from S=1 to T=8, the shortest path is (1, 3, 8). You can also go from S=1 to T=8  via (1, 2, 5, 8)  or (1, 4, 6, 7, 8) but these paths are not shortest.

You are given an array of integers 'ARR' of length 'N' and an integer Target. Your task is to return all pairs of elements such that they add up to Target.

Note:

We cannot use the element at a given index twice.

Follow Up:

Try to do this problem in O(N) time complexity. 

You have been given the preorder and inorder traversal of a binary tree. Your task is to construct a binary tree using the given inorder and preorder traversals.

Note:

You may assume that duplicates do not exist in the given traversals.

For example :

For the preorder sequence = [1, 2, 4, 7, 3] and the inorder sequence = [4, 2, 7, 1, 3], we get the following binary tree.

You have been given a complete binary tree of ‘N’ nodes. The nodes are numbered 1 to ‘N’.

You need to find the ‘next’ node that is immediately right in the level order form for each node in the given tree.

Note :

1. A complete binary tree is a binary tree in which nodes at all levels except the last level have two children and nodes at the last level have 0 children.
2. Node ‘U’ is said to be the next node of ‘V’ if and only if  ‘U’ is just next to ‘V’ in tree representation.
3. Particularly root node and rightmost nodes have ‘next’ node equal to ‘Null’ 
4. Each node of the binary tree has three-pointers, ‘left’, ‘right’, and ‘next’. Here ‘left’ and ‘right’ are the children of node and ‘next’ is one extra pointer that we need to update.

For Example :

The‘next’ node for ‘1’ is ‘Null’, ‘2’ has ‘next’ node ‘6’, ‘5’ has ‘next’ node ‘3’, For the rest of the nodes check below.

You have been given an array/list 'WORDS' of 'N' non-empty words, and an integer 'K'. Your task is to return the 'K' most frequent words sorted by their frequency from highest to lowest.

Note:

If two words have the same frequency then the lexicographically smallest word should come first in your answer.

Follow up:

Can you solve it in O(N * logK) time and O(N) extra space? 

What if the data size is too large and can not store in memory, ie. stored in a file and you are getting input as a stream. Given approach like dividing the file into smaller parts ( multi hashmap + heap)

 

10 mins discussion about current work .

You are given an unsorted array/list 'ARR' of 'N' integers. Your task is to return the length of the longest consecutive sequence.

The consecutive sequence is in the form ['NUM', 'NUM' + 1, 'NUM' + 2, ..., 'NUM' + L] where 'NUM' is the starting integer of the sequence and 'L' + 1 is the length of the sequence.

Note:

If there are any duplicates in the given array we will count only one of them in the consecutive sequence.

For example-

For the given 'ARR' [9,5,4,9,10,10,6].

Output = 3
The longest consecutive sequence is [4,5,6].

Follow Up:

Can you solve this in O(N) time and O(N) space complexity?

You’re given a board of N * M size. Each element of this board is represented by either 0 or 1, where 0 means ‘dead’ and 1 means ‘live’. Each element can interact with all of its eight neighbors using the following four rules.

If any live cell has less than two live neighbors, then it dies.

If any live cell has two or three live neighbors, then it lives onto the next generation.

If any live cell has more than three live neighbors, then it dies.

A dead element with exactly three live neighbors becomes a live element.

Your task is to print the next state of this board using these four rules.

Tell me an instance where your manager told you “ Dont do this”.

Design and implement a data structure for Least Recently Used (LRU) cache to support the following operations:

1. get(key) - Return the value of the key if the key exists in the cache, otherwise return -1.

2. put(key, value), Insert the value in the cache if the key is not already present or update the value of the given key if the key is already present. When the cache reaches its capacity, it should invalidate the least recently used item before inserting the new item.

You will be given ‘Q’ queries. Each query will belong to one of these two types:

Type 0: for get(key) operation.
Type 1: for put(key, value) operation.

Note :

1. The cache is initialized with a capacity (the maximum number of unique keys it can hold at a time).

2. Access to an item or key is defined as a get or a put operation on the key. The least recently used key is the one with the oldest access time.

Tell me the time when you entered into an argument with the team

A story-based question was there like A plant is there. It will drop two seeds one on left and one on right. Any one or both of them can convert to grown-up plants. Then again they will drop seeds left and right.
Finally boils down to a full binary tree is there. Any node is either plant or empty space (NULL). At any level length of pipes required to water the plant is the distance between the left most node ( grown-up plant ) - the rightmost node (grown-up plant).

The main focus was on how will you store the full binary tree representation. if you make a tree. Let's say n (number of levels) is too high. Eg 1000. How much memory it will consume. Is it feasible for a much larger n?

Finally, I took the approach. I will store it in a file in a serialized tree manner.
https://leetcode.com/problems/serialize-and-deserialize-binary-tree/
They will traverse level by level and calculate the value required for each level.