TIL - 250418_(SQL_datalemur_1개)

문제1 (링크)

WITH a AS (
  SELECT 
    *
    , SUM(num_users) OVER (ORDER BY searches) n_sum
    , SUM(num_users) OVER () tot
  FROM search_frequency
  ORDER BY searches
)
SELECT
  CASE WHEN tot/2 = n_sum THEN searches ELSE 0
FROM a