SDE - 2

Discussion about current work projects. Lengthy discussion on some technical issue that you were stuck for a long time. How did you find the issue, how did you fix it, could you do it any better now.

You are given a string “S”. Your task is to rearrange the characters of a string “S”, such that it does not contain any two adjacent characters which are the same.

If it is possible to rearrange the string “S”, then print any possible arrangement. else, print “not possible” without quotes.

For Example:

For a string  “qaacde”, This string has two same adjacent characters.
So, one possible way to rearrange the string is “qacade”. Now, this string does not have two adjacent characters that are the same. 

You are given an arbitrary binary tree, a node of the tree, and an integer 'K'. You need to find all such nodes which have a distance K from the given node and return the list of these nodes.

Distance between two nodes in a binary tree is defined as the number of connections/edges in the path between the two nodes.

Note:

1. A binary tree is a tree in which each node has at most two children. 
2. The given tree will be non-empty.
3. The given tree can have multiple nodes with the same value.
4. If there are no nodes in the tree which are at distance = K from the given node, return an empty list.
5. You can return the list of values of valid nodes in any order. For example if the valid nodes have values 1,2,3, then you can return {1,2,3} or {3,1,2} etc.

Example :

Consider this tree above. The target node is 5 and K = 3. The nodes at distance 1 from node 5 are {2}, nodes at distance 2 from node 5 are {1, 4} and nodes at distance 3 from node 5 are {6, 3}.

Given ‘N’ events each with a ‘start’ and ‘end’ time in the form of intervals such that it represents a booking on the half-open interval [start, end), i.e., the range of real numbers x such that start <= x < end. Initially, the calendar has no events. A new event can be added to the calendar, if adding the event will not cause a triple booking. A triple booking happens when three events have some non-empty intersection (i.e., there exists a time common to all the 3 events).

Your task is to process the 'N' events, and for each of the events, add that event to the calendar and return ‘True’, if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return ‘False’ and do not add that event to the calendar.

Given a System API to manage memory which allocates memory of size 1024 bytes only, create a wrapper to allocate memory efficiently. Deep discussion about why I used X data structure why not Y.

What keeps you motivated?

Who is your role model?