Rust

In this problem, I had to find two numbers that add up to 2020. While I could have put two loops, it's possible to reorder the terms in a + b = 2020 to 2020 - a = b. This allows retrieving b by merely knowing what a is, and this number could be checked for existence in a list in constant time with hash sets.

I also used this fact in second part of a puzzle which allowed to reduce the complexity from O(N^3​) to O(N^2​).