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• SQL
  
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  • № 1.
    
  • № 2.
    
  • № 3. ( )
    
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    • 3
  • № 4.
    
  • № 5.
    
  • № 6.
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  • № 1.
    
  • № 2. ( )
    
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    • 2
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  • № 1.
    
  • № 2. ( )
    
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  • № 3.

Hypothèses formulées et comment utiliser le manuel

• Les erreurs de syntaxe mineures importent peu lors d'un entretien avec un tableau blanc ou un bloc-notes. Mais ils peuvent distraire l'intervieweur, essayez donc idéalement de réduire leur nombre afin de concentrer toute votre attention sur la logique.
  
• Les réponses données ne sont pas nécessairement le seul moyen de résoudre chaque problème. N'hésitez pas à écrire des commentaires avec des solutions supplémentaires que vous pouvez ajouter à ce guide!

Conseils pour résoudre des tâches complexes dans les entretiens SQL

1. Écoutez attentivement la description du problème, répétez l'essence du problème à l'intervieweur
   
2. Formulez un cas limite pour démontrer que vous comprenez vraiment le problème (c'est-à-dire une ligne qui ne sera pas incluse dans la requête SQL finale que vous allez écrire)
   
3. ( ) , — : ,
   
   • , ,
4. SQL, , . , .

Tâches d'auto-association

No. 1. Variation en pourcentage d'un mois à l'autre

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

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' 

No. 2. Structure de l'arbre de marquage

parent de nœud
12
2 5
3 5
4 3
5 NUL 

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

No. 3. Rétention des utilisateurs par mois (plusieurs parties)

Partie 1

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

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

Partie 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)

3e partie

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

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

No. 4. Augmentation du total

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

| date | cumulative_cf |
| ------------ | --------------- |
| 01-01-2018 | -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

No. 5. Moyenne mobile

| date | inscriptions |
| ------------ | ---------- |
| 01-01-2018 | 10 |
| 2018-01-02 | 20 |
| 2018-01-03 | 50 |
| ... | ... |
| 01/10/2018 | 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

No. 6. Plusieurs conditions de connexion

| id | sujet | de | à | horodatage |
| ---- | ---------- | -------------- | -------------- | --- ------------------ |
| 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 | Running | 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 

Tâches pour les fonctions de fenêtre

Non. 1. Trouvez l'identifiant avec la valeur maximale

  depname | empno | salaire |     
----------- + ------- + -------- +
 développer | 11 | 5200 |
 développer | 7 | 4200 |
 développer | 9 | 4500 |
 développer | 8 | 6000 |
 développer | 10 | 5200 |
 personnel | 5 | 3500 |
 personnel | 2 | 3900 |
 ventes | 3 | 4800 |
 ventes | 1 | 5000 |
 ventes | 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;

No. 2. Valeur moyenne et classement avec une fonction de fenêtre (plusieurs parties)

Partie 1

  depname | empno | salaire |     
----------- + ------- + -------- +
 développer | 11 | 5200 |
 développer | 7 | 4200 |
 développer | 9 | 4500 |
 développer | 8 | 6000 |
 développer | 10 | 5200 |
 personnel | 5 | 3500 |
 personnel | 2 | 3900 |
 ventes | 3 | 4800 |
 ventes | 1 | 5000 |
 ventes | 4 | 4800 |

  depname | empno | salaire | avg_salary |     
----------- + ------- + -------- + ------------ +
 développer | 11 | 5200 | 5020 |
 développer | 7 | 4200 | 5020 |
 développer | 9 | 4500 | 5020 |
 développer | 8 | 6000 | 5020 |
 développer | 10 | 5200 | 5020 |
 personnel | 5 | 3500 | 3700 |
 personnel | 2 | 3900 | 3700 |
 ventes | 3 | 4800 | 4867 |
 ventes | 1 | 5000 | 4867 |
 ventes | 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

Partie 2

  depname | empno | salaire | salaire_rank |     
----------- + ------- + -------- + ------------- +
 développer | 11 | 5200 | 2 |
 développer | 7 | 4200 | 5 |
 développer | 9 | 4500 | 4 |
 développer | 8 | 6000 | 1 |
 développer | 10 | 5200 | 2 |
 personnel | 5 | 3500 | 2 |
 personnel | 2 | 3900 | 1 |
 ventes | 3 | 4800 | 2 |
 ventes | 1 | 5000 | 1 |
 ventes | 4 | 4800 | 2 |

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

Autres tâches de difficulté moyenne et élevée

N ° 1. Histogrammes

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

| seau | compter |
| --------- | ------- |
| 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 

No. 2. Connexion croisée (plusieurs parties)

Partie 1

| état | 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 

Partie 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 

No. 3. Calculs avancés

| utilisateur | classe |
| ------ | ------- |
| 1 | a |
| 1 | b |
| 1 | b |
| 2 | b |
| 3 | a |

| classe | compter |
| ------- | ------- |
| 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