This is a great idea, if I understood correctly this solution could be implemented as O(k*log(n) + k^2*n) = O(k^2*n) basically you have to do one sort for each index(meaning at most k being the length of the words) and each of those k times you can compare each of the strings with its next string, that taking O(k*n) k times and since we can assume k<<n this solves the problem. Brilliant! I will implement this ASAP.

## re: Advent of code Day 2 Part 2 Complexity VIEW POST

VIEW PARENT COMMENT VIEW FULL DISCUSSIONThis is a great idea, if I understood correctly this solution could be implemented as

`O(k*log(n) + k^2*n) = O(k^2*n)`

basically you have to do one sort for each index(meaning at most k being the length of the words) and each of those k times you can compare each of the strings with its next string, that taking`O(k*n)`

k times and since we can assume`k<<n`

this solves the problem. Brilliant! I will implement this ASAP.