题目：原题链接（简单）
标签：SQL
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | 902ms (57.14%) | ||
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
SELECT customer_id,
COUNT(DISTINCT visit_id) AS count_no_trans
FROM Visits
WHERE visit_id NOT IN (SELECT visit_id FROM Transactions)
GROUP BY customer_id;
