Senior Java Developer

You are given a list of ‘N’ non-overlapping intervals (each interval can be represented using two integers ‘start’ and ‘end’), sorted in ascending order (based on ‘start’ values). Your task is to insert a given interval in the list, such that all the intervals are present in sorted order.

In case the given interval overlaps with other intervals, merge all necessary intervals to produce a list containing only non-overlapping intervals.

Example:

Let’s say the list of intervals is: [[1,3], [5,7], [8,12]] and we need to insert the interval [4,6] into the list. [4,6] must be inserted in the second position. After insertion, since [4,6] overlaps with [5,7], we merge them into one interval [4,7]. So, our resulting list is:  [[1,3], [4,7], [8,12]]

You are given a circular array 'a' of length 'n'.

A circular array is an array in which we consider the first element is next of the last element. That is, the next element of 'a[n - 1]' is 'a[0]'.

Find the Next Greater Element(NGE) for every element.

The Next Greater Element for an element 'x' is the first element on the right side of 'x' in the array, which is greater than 'x'.

If no greater elements exist to the right of 'x', consider the next greater element as -1.

Example:

Input: 'a' = [1, 5, 3, 4, 2]

Output: NGE = [5, -1, 4, 5, 5]

Explanation: For the given array,

- The next greater element for 1 is 5.

- There is no greater element for 5 on the right side. So we consider NGE as -1.

- The next greater element for 3 is 4.

- The next greater element for 4 is 5, when we consider the next elements as 4 -> 2 -> 1 -> 5.

- The next greater element for 2 is 5, when we consider the next elements as 2 -> 1 -> 5.

Design a data structure that satisfies the constraints of a Least Recently Used (LRU).

1. Get(int num): If the key exists, it will return the value of the key stored. Else, return -1.    
2. Put(int key, int value): If the key already exists, update the value of the key. Else add the key-value pair to the cache. If the number of keys is more than the capacity for this operation, delete the least recently key used. 

Example:

For the following input: 

4 2
2 1 4
1 1
1 4

We will initialize an empty LRU cache for the first operation with a maximum capacity of 2.
For the first operation, we need to add a key-value pair (1,4) to the cache.
For the second operation, we need to return the value stored for key 1, i.e., 4
For the third operation, we need to return -1, as we don't have any key 4 in the cache.

So, the final output will be: 
4  -1

1. Discussion about the company's future goals and anything you wanna discuss about the business and its scope.

2. Tell me about your achievements.

3. Why do you want to join Zypp electric?

4. What are the exciting features that you have worked on?