# Posts

• ## Leetcode: Find peak element

Given an array `n` of integers, find one peak in `O(log*n)` time. A peak is defined as a number where the two numbers immediately left and right are strictly less than the number. Numbers outside of the bounds of this array are considered to be smaller.

• ## Leetcode: Minimum Stack

Create a stack implementation that is able to return the minimum element as well as `push` and `pop` elements, all in constant time `O(1)`.

• ## Leetcode: Populate next pointers in tree

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.

• ## Leetcode: Generate parentheses

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 `1` and `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 `(())` and `()()` etc. 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 `()()()` or `()(())` (for `n=3`) because they would already been visited.

• ## Leetcode: Word break

Given a string `s` and a list of `words`, return `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.

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