SDE - 1

You have a binary tree of 'N' unique nodes and a Start node from where the tree will start to burn. Given that the Start node will always exist in the tree, your task is to print the time (in minutes) that it will take to burn the whole tree.

It is given that it takes 1 minute for the fire to travel from the burning node to its adjacent node and burn down the adjacent node.

For Example :

For the given binary tree: [1, 2, 3, -1, -1, 4, 5, -1, -1, -1, -1]
Start Node: 3

    1
   / \
  2   3
     / \
    4   5

Output: 2

Explanation :
In the zeroth minute, Node 3 will start to burn.

After one minute, Nodes (1, 4, 5) that are adjacent to 3 will burn completely.

After two minutes, the only remaining Node 2 will be burnt and there will be no nodes remaining in the binary tree. 

So, the whole tree will burn in 2 minutes.

Design and implement a Least Frequently Used(LFU) Cache, to implement the following functions:

1. put(U__ID, value): Insert the value in the cache if the key(‘U__ID’) 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 frequently used item before inserting the new item.

2. get(U__ID): Return the value of the key(‘U__ID’),  present in the cache, if it’s present otherwise return -1.

Note:

  1) The frequency of use of an element is calculated by a number of operations with its ‘U_ID’ performed after it is inserted in the cache.

  2) If multiple elements have the least frequency then we remove the element which was least recently used. 

You have been given ‘M’ operations which you need to perform in the cache. Your task is to implement all the functions of the LFU cache.

Type 1: for put(key, value) operation.
Type 2: for get(key) operation.

Example:

We perform the following operations on an empty cache which has capacity 2:

When operation 1 2 3 is performed, the element with 'U_ID' 2 and value 3 is inserted in the cache.

When operation 1 2 1 is performed, the element with 'U_ID' 2’s value is updated to 1.  

When operation 2 2 is performed then the value of 'U_ID' 2 is returned i.e. 1.

When operation 2 1 is performed then the value of 'U_ID' 1 is to be returned but it is not present in cache therefore -1 is returned.

When operation 1 1 5 is performed, the element with 'U_ID' 1 and value 5 is inserted in the cache. 

When operation 1 6 4 is performed, the cache is full so we need to delete an element. First, we check the number of times each element is used. Element with 'U_ID' 2 is used 3 times (2 times operation of type 1 and 1-time operation of type 1). Element with 'U_ID' 1 is used 1 time (1-time operation of type 1). So element with 'U_ID' 1 is deleted. The element with 'U_ID' 6 and value 4 is inserted in the cache. 

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