DEV Community

Discussion on: Advent of Code 2020 Solution Megathread - Day 2: Password Philosophy

View post
Collapse
 
clothierdroid profile image
David Clothier • • Edited

SQL (0 lines of code huehue)

Get this table naming 'day2'

table

2.1. Solution

SELECT COUNT(*) FROM 
(
SELECT
   letra
 , pass
 , nmin
 , nmax
 ,(LENGTH(pass) - LENGTH(REPLACE(pass, letra, ''))) AS times
FROM
   day2
HAVING
   times BETWEEN nmin AND nmax
) puzzle_answer
Enter fullscreen mode Exit fullscreen mode

2.2. Solution

SELECT COUNT(*) 
FROM   day2 
WHERE  LOCATE(letra, pass, nmin) = nmin
       XOR LOCATE(letra, pass, nmax) = nmax
Enter fullscreen mode Exit fullscreen mode