看看你离世界一流大厂有多远？3道Google最新SQL面试题 ⛵
原创看看你离世界一流大厂有多远？3道Google最新SQL面试题 ⛵
原创
ShowMeAI
发布于 2022-08-09 20:52:12
发布于 2022-08-09 20:52:12
💡 作者：韩信子@ShowMeAI 📘 数据分析◉技能提升系列：https://www.showmeai.tech/tutorials/33 📘 AI 面试题库系列：https://www.showmeai.tech/tutorials/48 📘 本文地址：https://www.showmeai.tech/article-detail/297 📢 声明：版权所有，转载请联系平台与作者并注明出处 📢 收藏ShowMeAI查看更多精彩内容
下面是最新的 3 道 Google SQL 面试题和参考答案。这些题目面向的 Google 职位包括：数据科学 家、数据分析师、商业智能 工程师、数据工程师和商业分析师。
ShowMeAI 制作了快捷即查即用的 SQL 速查表手册，大家可以在下述位置获得：编程语言速查表 | SQL 速查表
💡 面试题 1：墨西哥和美国第三高峰
问题： 请完成1个 SQL 来找出每个国家第三高的山名，并按 ASC 顺序对国家/地区排序。
Table: mountains
+---------------------+------+-------------+
|name |height|country |
+---------------------+------+-------------+
|Denalli |20310 |United States|
|Saint Elias |18008 |United States|
|Foraker |17402 |United States|
|Pico de Orizab |18491 |Mexico |
|Popocatépetl |17820 |Mexico |
|Iztaccihuatl |17160 |Mexico |
+---------------------+------+-------------+参考答案：
SELECT "country",
"name"
FROM (SELECT "country",
"name",
Rank()
OVER (
partition BY "country"
ORDER BY "height" DESC) AS "rank"
FROM mountains) AS m
WHERE "rank" = 3
ORDER BY country ASC 💡 面试题 2：用 latest_event 查找当前打开的页数
问题： 给定下表，表中包含有关页面状态更改时间的信息。完成 SQL 查找当前使用 latest_event 的页面数。 注意，表中 page_flag 列将用于识别页面是『OFF』还是『ON』。
Table: pages_info
+-------+--------------------------------------+----------+
|page_id|event_time |page_flag |
+-------+--------------------------------------+----------+
|1 |current_timestamp - interval '6 hours'|ON |
|1 |current_timestamp - interval '3 hours'|OFF |
|1 |current_timestamp - interval '1 hours'|ON |
|2 |current_timestamp - interval '3 hours'|ON |
|2 |current_timestamp - interval '1 hours'|OFF |
|3 |current_timestamp |ON |
+-------+--------------------------------------+----------+参考答案：
-- 首先，对于每个页面ID，让我们选择最新的记录（基于事件时间列）。
SELECT page_id,
Max(event_time) AS latest_event
FROM pages_info
GROUP BY page_id
-- 接着，我们将前面的查询与原表连接起来，并检查其中有多少人的标记页等于ON。
WITH latest_event
AS (SELECT page_id,
Max(event_time) AS latest_event
FROM pages_info
GROUP BY page_id)
SELECT Sum(CASE
WHEN page_flag = 'ON' THEN 1
ELSE 0
END) AS result
FROM pages_info pi
JOIN latest_event le
ON pi.page_id = le.page_id
AND pi.event_time = le.latest_event; 💡 面试题 3：回访用户
问题： 在如下的数据库表中，包含有关用户访问网页的信息。 完成 SQL 返回连续访问该页面最长的 3 个用户，按长短的倒序排列 3 个用户。
Table: visits
+--------+----------------------------+
|user_id |date |
+--------+----------------------------+
|1 |current_timestamp::DATE - 0 |
|1 |current_timestamp::DATE - 1 |
|1 |current_timestamp::DATE - 2 |
|1 |current_timestamp::DATE - 3 |
|1 |current_timestamp::DATE - 4 |
|2 |current_timestamp::DATE - 1 |
|4 |current_timestamp::DATE - 0 |
|4 |current_timestamp::DATE - 1 |
|4 |current_timestamp::DATE - 3 |
|4 |current_timestamp::DATE - 4 |
|4 |current_timestamp::DATE - 62|
+--------+----------------------------+参考答案：
--首先，让我们添加一个新的列，其值是每个用户的下一次访问（与当前日期不同）。我们将使用lead函数来完成:
SELECT DISTINCT user_id,
date,
Lead(date)
OVER (
partition BY user_id
ORDER BY date) AS next_date
FROM (SELECT DISTINCT *
FROM visits) AS t;
--接着，让我们创建另一个列，其目的是让我们知道访问的停止。这包括检查下一个日期是否与当前日期+1是否不同。
WITH next_dates
AS (SELECT DISTINCT user_id,
date,
Lead(date)
OVER (
partition BY user_id
ORDER BY date) AS next_date
FROM (SELECT DISTINCT *
FROM visits) AS t) --去重
SELECT user_id,
date,
next_date,
CASE
WHEN next_date IS NULL
OR next_date = date + 1 THEN 1
ELSE NULL
END AS streak
FROM next_dates;
--接着，我们将为每个用户创建一个分区，每个分区代表一个连续的访问。从概念上讲，我们要做的是，对于每个用户，取最近的记录（基于日期）并赋值为0，然后寻找下面的记录，如果访问没有停止就赋值为0，如果访问停止就赋值为1（如果连胜列为空），然后继续这样做，直到每个连续访问被一个不同的分区所代表。执行这一逻辑的代码如下。
WITH next_dates
AS (SELECT DISTINCT user_id,
date,
Lead(date)
OVER (
partition BY user_id
ORDER BY date) AS next_date
FROM (SELECT DISTINCT *
FROM visits)),
streaks
AS (SELECT user_id,
date,
next_date,
CASE
WHEN next_date IS NULL
OR next_date = date + 1 THEN 1
ELSE NULL
END AS streak
FROM next_dates)
SELECT *,
Sum(CASE
WHEN streak IS NULL THEN 1
ELSE 0
END)
OVER (
partition BY user_id
ORDER BY date) AS partition
FROM streaks;
--一旦我们有了这个分区，问题就容易了，现在我们只需要计算每个用户和分区的记录数，并找到计数最多的用户。完整的查询如下
WITH next_dates AS
(
SELECT DISTINCT user_id,
date,
Lead(date) OVER (partition BY user_id ORDER BY date) AS next_date
FROM visits ), streaks AS
(
SELECT user_id,
date,
next_date,
CASE
WHEN next_date IS NULL
OR next_date = date + 1 THEN 1
ELSE NULL
END AS streak
FROM next_dates ), partitions AS
(
SELECT *,
Sum(
CASE
WHEN streak IS NULL THEN 1
ELSE 0
END ) OVER (partition BY user_id ORDER BY date) AS partition
FROM streaks ), count_partitions AS
(
SELECT user_id,
partition,
Count(1) AS streak_days
FROM partitions
GROUP BY user_id,
partition )
SELECT user_id,
Max(streak_days) AS longest_streak
FROM count_partitions
GROUP BY user_id
ORDER BY 2 DESC limit 3;参考资料
- 📘 编程语言速查表 | SQL 速查表：https://www.showmeai.tech/article-detail/99
原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。
如有侵权，请联系 cloudcommunity@tencent.com 删除。
原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。
如有侵权，请联系 cloudcommunity@tencent.com 删除。
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