SDE - 1

You have been given an array 'PRICES' consisting of 'N' integers where PRICES[i] denotes the price of a given stock on the i-th day. You are also given an integer 'K' denoting the number of possible transactions you can make.

Your task is to find the maximum profit in at most K transactions. A valid transaction involves buying a stock and then selling it.

Note

You can’t engage in multiple transactions simultaneously, i.e. you must sell the stock before rebuying it.

For Example

Input: N = 6 , PRICES = [3, 2, 6, 5, 0, 3] and K = 2.
Output: 7

Explanation : The optimal way to get maximum profit is to buy the stock on day 2(price = 2) and sell it on day 3(price = 6) and rebuy it on day 5(price = 0) and sell it on day 6(price = 3). The maximum profit will be (6 - 2) + (3 - 0) = 7.

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. 

You have been given an array 'ARR' of ‘N’ integers. You have to return the minimum number of jumps needed to reach the last index of the array i.e ‘N - 1’.

From index ‘i’, we can jump to an index ‘i + k’ such that 1<= ‘k’ <= ARR[i] .

'ARR[i]' represents the maximum distance you can jump from the current index.

If it is not possible to reach the last index, return -1.

Note:

Consider 0-based indexing.

Example:

Consider the array 1, 2, 3, 4, 5, 6 
We can Jump from index 0 to index 1
Then we jump from index 1 to index 2
Then finally make a jump of 3 to reach index N-1

There is also another path where
We can Jump from index 0 to index 1
Then we jump from index 1 to index 3
Then finally make a jump of 2 to reach index N-1

So multiple paths may exist but we need to return the minimum number of jumps in a path to end which here is 3.

Design a Web Crawler.