Wow, 5 minutes in Ruby, and less than a second in Go? I know scripting languages are slow, but it's easy to forget just how much slower. (EDIT: Corrected misrea

Wow, 5 minutes in Ruby, and less than a second in Go? I know scripting languages are slow, but it's easy to forget just how much slower. (EDIT: Corrected misread number thanks to dljsjr). I don't want to detract from the article's main point with a cliche "algorithmic optimization beats fine tuning", but I think it's worth mentioning. Unless I'm mistaken, this particular problem can be solved about 10,000,000x more efficiently. > Given two numbers X and Y, return how many numbers in that range > (inclusive) are palindromes, and also the square of a palindrome. I would just count the digits of X and Y, then ranging between those counts, apply the following algorithm: For N digits, enumerate all numbers of N/2 digits satisfying the bounds X and Y. If N is even then simply mirror the first N/2 digits. If N is odd, mirror the first N/2-1. Finally, check the bounds of the resulting number against X and Y to filter out edge cases. If the original algorithm runs in 10^N time, this would be 10^(N/2). For N=14 as mentioned, that's roughly a speedup of 10,000,000. Edit: It seems I misinterpreted the question (which sounds like you're supposed to enumerate all palindrome numbers X>=N>=Y, and also print out the squares of those numbers). This is actually asking for a particular type of palindromes -- those which are palindromic squares of palindromes. As macobo helpfully points out, you can drastically optimize mathematically from this additional filter.