Given a perfect binary search tree, connect all the nodes to the node on their right The solution is to do breadth-first-search in an iterative fashion. This puts all the nodes at the same level in a queue. By iterating over the queue we can connect each node to its successor. During this process we add the children of each node as the next entry. We repeat this process until the queue is empty.
Given a strictly positive integer
n, write a function that returns all possible combinations of well-formed parentheses.
Parentheses can be nested and added one after the other. It is important that we don’t create invalid combinations, such as
)(. The idea then becomes to start with a single set of parentheses
(). We can add another set of parentheses at three possible places:
1(2)3. When looking closely, we see that
3 are the same position.
We can then utilize Pythons string splitting capabilities which allow us to insert one or more characters at any place in the string. We do this by iterating over the string and inserting
() at every possible position. This creates all valid pairs like
To avoid the aforementioned duplicates we can add a memory to the function and store all the visited possible combinations. This allows us to speed the process up significantly. For example when we visit
()(), we don’t need to visit it again to form
n=3) because they would already been visited.
Given a string
s and a list of
True if the string can be constructed from any combination of the words and
False otherwise. The alphabet contains only lowercase English characters.
My initial idea was to replace all occurrences of a
word in the string
s. The problem with this approach is that a string
aabb with the words
['ab'] is considered valid, while it is not. I then tried on adding breaking characters (
.) to prevent this. It worked although very slowly.
Given an array
nums with integer values
-10 <= x <= 10, calculate the maximum consecutive product. For example, the array
[2,3,-1,4] produces the maximum product
The first obsveration that we can make is that whenever a
0 is encountered, it sets the whole product to
0. This means, if a
0 is somewhere in the middle of the array, we need to look at the left and right part individually because they cannot be connected.
Secondly, an odd amount of negative numbers makes the whole product negative. Having
6, … negative numbers will keep the product at the same amount (if all of them are
-1) or increase it.
Finally, since those numbers are all integers, the longer the chain of multiplied numbers, the higher to outcome.
Given a single linked list, sort the values in ascending order.