Inhalt

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Annahmen getroffen und wie man das Handbuch verwendet

• Kleinere Syntaxfehler spielen bei einem Interview mit einem Whiteboard oder Notizblock keine große Rolle. Aber sie können den Interviewer ablenken, also versuchen Sie im Idealfall, ihre Anzahl zu reduzieren, um alle Aufmerksamkeit auf die Logik zu konzentrieren.
  
• Die gegebenen Antworten sind nicht unbedingt die einzige Möglichkeit, jedes Problem zu lösen. Fühlen Sie sich frei, Kommentare mit zusätzlichen Lösungen zu schreiben, die Sie diesem Handbuch hinzufügen können!

Tipps zum Lösen komplexer Aufgaben in SQL-Interviews

1. Hören Sie sich die Beschreibung des Problems genau an und wiederholen Sie dem Interviewer die Essenz des Problems
   
2. Formulieren Sie einen Grenzfall, um zu demonstrieren, dass Sie das Problem wirklich verstehen (d. H. Eine Zeile, die nicht in der endgültigen SQL-Abfrage enthalten ist, die Sie schreiben werden).
   
3. ( ) , — : ,
   
   • , ,
4. SQL, , . , .

Selbstassoziationsaufgaben

Nr. 1. Prozentuale Änderung von Monat zu Monat

| user_id | Datum |
| --------- | ------------ |
| 1 | 2018-07-01 |
| 234 | 2018-07-02 |
| 3 | 2018-07-02 |
| 1 | 2018-07-02 |
| ... | ... |
| 234 | 2018-10-04 |

WITH mau AS 
(
  SELECT 
   /* 
    *       
    *  , . .   ,    . 
    *    ,   
    *
    *  Postgres  DATE_TRUNC(),   
    *      SQL   
    * . https://www.postgresql.org/docs/9.0/functions-datetime.html#FUNCTIONS-DATETIME-TRUNC
    */ 
    DATE_TRUNC('month', date) month_timestamp,
    COUNT(DISTINCT user_id) mau
  FROM 
    logins 
  GROUP BY 
    DATE_TRUNC('month', date)
  )
 
 SELECT 
    /*
    *    SELECT      . 
    * 
    *        ,   
    *    ,   , 
    *        ..
    */ 
    a.month_timestamp previous_month, 
    a.mau previous_mau, 
    b.month_timestamp current_month, 
    b.mau current_mau, 
    ROUND(100.0*(b.mau - a.mau)/a.mau,2) AS percent_change 
 FROM
    mau a 
 JOIN 
    /*
    *   `ON b.month_timestamp = a.month_timestamp + interval '1 month'` 
    */
    mau b ON a.month_timestamp = b.month_timestamp - interval '1 month' 

Nr. 2. Baumstruktur markieren

Knoten übergeordnet
12
2 5
3 5
4 3
5 NULL 

node    label  
1       Leaf
2       Inner
3       Inner
4       Leaf
5       Root

WITH join_table AS
(
    SELECT 
        cur.node, 
        cur.parent, 
        COUNT(next.node) AS num_children
    FROM 
        tree cur
    LEFT JOIN 
        tree next ON (next.parent = cur.node)
    GROUP BY 
        cur.node, 
        cur.parent
)

SELECT
    node,
    CASE
        WHEN parent IS NULL THEN "Root"
        WHEN num_children = 0 THEN "Leaf"
        ELSE "Inner"
    END AS label
FROM 
    join_table 

SELECT 
    node,
    CASE 
        WHEN parent IS NULL THEN 'Root'
        WHEN node NOT IN 
            (SELECT parent FROM tree WHERE parent IS NOT NULL) THEN 'Leaf'
        WHEN node IN (SELECT parent FROM tree) AND parent IS NOT NULL THEN 'Inner'
    END AS label 
 from 
    tree

Nr. 3. Benutzerbindung pro Monat (mehrere Teile)

Teil 1

| user_id | Datum |
| --------- | ------------ |
| 1 | 2018-07-01 |
| 234 | 2018-07-02 |
| 3 | 2018-07-02 |
| 1 | 2018-07-02 |
| ... | ... |
| 234 | 2018-10-04 |

SELECT 
    DATE_TRUNC('month', a.date) month_timestamp, 
    COUNT(DISTINCT a.user_id) retained_users 
 FROM 
    logins a 
 JOIN 
    logins b ON a.user_id = b.user_id 
        AND DATE_TRUNC('month', a.date) = DATE_TRUNC('month', b.date) + 
                                             interval '1 month'
 GROUP BY 
    date_trunc('month', a.date)

WITH DistinctMonthlyUsers AS (
  /*
  *     ** , 
  *  
  */
    SELECT DISTINCT
      DATE_TRUNC('MONTH', a.date) AS month_timestamp,
      user_id
    FROM logins
  )

SELECT
  CurrentMonth.month_timestamp month_timestamp,
  COUNT(PriorMonth.user_id) AS retained_user_count
FROM 
    DistinctMonthlyUsers AS CurrentMonth
LEFT JOIN 
    DistinctMonthlyUsers AS PriorMonth
  ON
    CurrentMonth.month_timestamp = PriorMonth.month_timestamp + INTERVAL '1 MONTH'
    AND 
    CurrentMonth.user_id = PriorMonth.user_id

Teil 2

SELECT 
    DATE_TRUNC('month', a.date) month_timestamp, 
    COUNT(DISTINCT b.user_id) churned_users 
FROM 
    logins a 
FULL OUTER JOIN 
    logins b ON a.user_id = b.user_id 
        AND DATE_TRUNC('month', a.date) = DATE_TRUNC('month', b.date) + 
                                         interval '1 month'
WHERE 
    a.user_id IS NULL 
GROUP BY 
    DATE_TRUNC('month', a.date)

Teil 3

| user_id | month_date |
| --------- | ------------ |
| 1 | 2018-05-01 |
| 234 | 2018-05-01 |
| 3 | 2018-05-01 |
| 12 | 2018-05-01 |
| ... | ... |
| 234 | 2018-10-01 |

WITH user_login_data AS 
(
    SELECT 
        DATE_TRUNC('month', a.date) month_timestamp,
        a.user_id,
        /* 
        *   ,    SQL,   , 
        *      SELECT   HAVING.
        *       .  
        */ 
        MAX(b.month_date) as most_recent_churn, 
        MAX(DATE_TRUNC('month', c.date)) as most_recent_active 
     FROM 
        logins a
     JOIN 
        user_churns b 
            ON a.user_id = b.user_id AND DATE_TRUNC('month', a.date) > b.month_date 
     JOIN
        logins c 
            ON a.user_id = c.user_id 
            AND 
            DATE_TRUNC('month', a.date) > DATE_TRUNC('month', c.date)
     WHERE 
        DATE_TRUNC('month', a.date) = DATE_TRUNC('month', current_timestamp)
     GROUP BY 
        DATE_TRUNC('month', a.date),
        a.user_id
     HAVING 
        most_recent_churn > most_recent_active

Nr. 4. Zunehmende Summe

| Datum | cash_flow |
| ------------ | ----------- |
| 2018-01-01 | -1000 |
| 2018-01-02 | -100 |
| 2018-01-03 | 50 |
| ... | ... |

| Datum | cumulative_cf |
| ------------ | --------------- |
| 2018-01-01 | -1000 |
| 2018-01-02 | -1100 |
| 2018-01-03 | -1050 |
| ... | ... |

SELECT 
    a.date date, 
    SUM(b.cash_flow) as cumulative_cf 
FROM
    transactions a
JOIN b 
    transactions b ON a.date >= b.date 
GROUP BY 
    a.date 
ORDER BY 
    date ASC

SELECT 
    date, 
    SUM(cash_flow) OVER (ORDER BY date ASC) as cumulative_cf 
FROM
    transactions 
ORDER BY 
    date ASC

Nr. 5. Gleitender Durchschnitt

| Datum | sign_ups |
| ------------ | ---------- |
| 2018-01-01 | 10 |
| 2018-01-02 | 20 |
| 2018-01-03 | 50 |
| ... | ... |
| 2018-10-01 | 35 |

SELECT 
  a.date, 
  AVG(b.sign_ups) average_sign_ups 
FROM 
  signups a 
JOIN 
  signups b ON a.date <= b.date + interval '6 days' AND a.date >= b.date
GROUP BY 
  a.date

Nr. 6. Mehrere Verbindungsbedingungen

| id | Betreff | von | zu | Zeitstempel |
| ---- | ---------- | -------------- | -------------- | --- ------------------ |
| 1 | Yosemite | zach@g.com | thomas@g.com | 2018-01-02 12:45:03 |
| 2 | Big Sur | sarah@g.com | thomas@g.com | 2018-01-02 16:30:01 |
| 3 | Yosemite | thomas@g.com | zach@g.com | 2018-01-02 16:35:04 |
| 4 | Laufen | jill@g.com | zach@g.com | 2018-01-03 08:12:45 |
| 5 | Yosemite | zach@g.com | thomas@g.com | 2018-01-03 14:02:01 |
| 6 | Yosemite | thomas@g.com | zach@g.com | 2018-01-03 15:01:05 |
| .. | .. | .. | .. | .. |

SELECT 
    a.id, 
    MIN(b.timestamp) - a.timestamp as time_to_respond 
FROM 
    emails a 
JOIN
    emails b 
        ON 
            b.subject = a.subject 
        AND 
            a.to = b.from
        AND 
            a.from = b.to 
        AND 
            a.timestamp < b.timestamp 
 WHERE 
    a.to = 'zach@g.com' 
 GROUP BY 
    a.id 

Aufgaben für Fensterfunktionen

Nr. 1. Suchen Sie die Kennung mit dem Maximalwert

  depname | empno | Gehalt |     
----------- + ------- + -------- +
 entwickeln | 11 | 5200 |
 entwickeln | 7 | 4200 |
 entwickeln | 9 | 4500 |
 entwickeln | 8 | 6000 |
 entwickeln | 10 | 5200 |
 Personal | 5 | 3500 |
 Personal | 2 | 3900 |
 Vertrieb | 3 | 4800 |
 Vertrieb | 1 | 5000 |
 Vertrieb | 4 | 4800 |

WITH max_salary AS (
    SELECT 
        MAX(salary) max_salary
    FROM 
        salaries
    )
SELECT 
    s.empno
FROM 
    salaries s
JOIN 
    max_salary ms ON s.salary = ms.max_salary

WITH sal_rank AS 
  (SELECT 
    empno, 
    RANK() OVER(ORDER BY salary DESC) rnk
  FROM 
    salaries)
SELECT 
  empno
FROM
  sal_rank
WHERE 
  rnk = 1;

Nr. 2. Mittelwert und Rangfolge mit einer Fensterfunktion (mehrere Teile)

Teil 1

  depname | empno | Gehalt |     
----------- + ------- + -------- +
 entwickeln | 11 | 5200 |
 entwickeln | 7 | 4200 |
 entwickeln | 9 | 4500 |
 entwickeln | 8 | 6000 |
 entwickeln | 10 | 5200 |
 Personal | 5 | 3500 |
 Personal | 2 | 3900 |
 Vertrieb | 3 | 4800 |
 Vertrieb | 1 | 5000 |
 Vertrieb | 4 | 4800 |

  depname | empno | Gehalt | avg_salary |     
----------- + ------- + -------- + ------------ +
 entwickeln | 11 | 5200 | 5020 |
 entwickeln | 7 | 4200 | 5020 |
 entwickeln | 9 | 4500 | 5020 |
 entwickeln | 8 | 6000 | 5020 |
 entwickeln | 10 | 5200 | 5020 |
 Personal | 5 | 3500 | 3700 |
 Personal | 2 | 3900 | 3700 |
 Vertrieb | 3 | 4800 | 4867 |
 Vertrieb | 1 | 5000 | 4867 |
 Vertrieb | 4 | 4800 | 4867 |

SELECT 
    *, 
    /*
    * AVG() is a Postgres command, but other SQL flavors like BigQuery use 
    * AVERAGE()
    */ 
    ROUND(AVG(salary),0) OVER (PARTITION BY depname) avg_salary
FROM
    salaries

Teil 2

  depname | empno | Gehalt | Gehalt_Rang |     
----------- + ------- + -------- + ------------- +
 entwickeln | 11 | 5200 | 2 |
 entwickeln | 7 | 4200 | 5 |
 entwickeln | 9 | 4500 | 4 |
 entwickeln | 8 | 6000 | 1 |
 entwickeln | 10 | 5200 | 2 |
 Personal | 5 | 3500 | 2 |
 Personal | 2 | 3900 | 1 |
 Vertrieb | 3 | 4800 | 2 |
 Vertrieb | 1 | 5000 | 1 |
 Vertrieb | 4 | 4800 | 2 |

SELECT 
    *, 
    RANK() OVER(PARTITION BY depname ORDER BY salary DESC) salary_rank
 FROM  
    salaries 

Andere Aufgaben mittlerer und hoher Schwierigkeit

Nr. 1. Histogramme

| session_id | length_seconds |
| ------------ | ---------------- |
| 1 | 23 |
| 2 | 453 |
| 3 | 27 |
| .. | .. |

| Eimer | count |
| --------- | ------- |
| 20-25 | 2 |
| 450-455 | 1 |

WITH bin_label AS 
(SELECT 
    session_id, 
    FLOOR(length_seconds/5) as bin_label 
 FROM
    sessions 
 )
 SELECT 
    CONCATENTATE(STR(bin_label*5), '-', STR(bin_label*5+5)) bucket, 
    COUNT(DISTINCT session_id) count 
 GROUP BY 
    bin_label
 ORDER BY 
    bin_label ASC 

Nr. 2. Querverbindung (mehrere Teile)

Teil 1

| Zustand | total_streams |
| ------- | --------------- |
| NC | 34569 |
| SC | 33999 |
| CA | 98324 |
| MA | 19345 |
| .. | .. |

| state_a | state_b |
| --------- | --------- |
| NC | SC |
| SC | NC |

SELECT
    a.state as state_a, 
    b.state as state_b 
 FROM   
    state_streams a
 CROSS JOIN 
    state_streams b 
 WHERE 
    ABS(a.total_streams - b.total_streams) < 1000
    AND 
    a.state <> b.state 

SELECT
    a.state as state_a, 
    b.state as state_b 
 FROM   
    state_streams a, state_streams b 
 WHERE 
    ABS(a.total_streams - b.total_streams) < 1000
    AND 
    a.state <> b.state 

Teil 2

SELECT
    a.state as state_a, 
    b.state as state_b 
 FROM   
    state_streams a, state_streams b 
 WHERE 
    ABS(a.total_streams - b.total_streams) < 1000
    AND 
    a.state > b.state 

Nr. 3. Erweiterte Berechnungen

| Benutzer | Klasse |
| ------ | ------- |
| 1 | a |
| 1 | b |
| 1 | b |
| 2 | b |
| 3 | a |

| Klasse | count |
| ------- | ------- |
| a | 1 |
| b | 2 |

WITH usr_b_sum AS 
(
    SELECT 
        user, 
        SUM(CASE WHEN class = 'b' THEN 1 ELSE 0 END) num_b
    FROM 
        table
    GROUP BY 
        user
), 

usr_class_label AS 
(
    SELECT 
        user, 
        CASE WHEN num_b > 0 THEN 'b' ELSE 'a' END class 
    FROM 
        usr_b_sum
)

SELECT 
    class, 
    COUNT(DISTINCT user) count 
FROM
    usr_class_label
GROUP BY 
    class 
ORDER BY 
    class ASC

SELECT 
    "a" class,
    COUNT(DISTINCT user_id) - 
        (SELECT COUNT(DISTINCT user_id) FROM table WHERE class = 'b') count 
UNION
SELECT 
    "b" class,
    (SELECT COUNT(DISTINCT user_id) FROM table WHERE class = 'b') count