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Suposições feitas e como usar o manual

• Pequenos erros de sintaxe não importam muito durante uma entrevista com um quadro branco ou bloco de notas. Mas eles podem distrair o entrevistador, então, idealmente, tente reduzir seu número para concentrar toda a atenção na lógica.
  
• As respostas dadas não são necessariamente a única maneira de resolver cada problema. Sinta-se livre para escrever comentários com soluções adicionais que você pode adicionar a este guia!

Dicas para resolver tarefas complexas em entrevistas SQL

1. Ouça atentamente a descrição do problema, repita a essência do problema para o entrevistador.
   
2. Formule um caso limite para demonstrar que realmente entende o problema (ou seja, uma linha que não será incluída na consulta SQL final que você escreverá)
   
3. ( ) , — : ,
   
   • , ,
4. SQL, , . , .

Tarefas de auto-associação

No. 1. Variação percentual mês a mês

| user_id | data |
| --------- | ------------ |
| 1 | 01/07/2018 |
| 234 02/07/2018 |
| 3 02/07/2018 |
| 1 | 02/07/2018 |
| ... ...
| 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. Estrutura da árvore de marcação

pai do nó
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

No. 3. Retenção de usuários por mês (várias partes)

Parte 1

| user_id | data |
| --------- | ------------ |
| 1 | 01/07/2018 |
| 234 02/07/2018 |
| 3 02/07/2018 |
| 1 | 02/07/2018 |
| ... ...
| 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

Parte 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)

Parte 3

| user_id | month_date |
| --------- | ------------ |
| 1 | 01/05/2018 |
| 234 01/05/2018 |
| 3 01/05/2018 |
| 12 01/05/2018 |
| ... ...
| 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. Total crescente

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

| data | cumulative_cf |
| ------------ | --------------- |
| 01-01-2018 | -1000
| 02-01-2018 | -1100
| 03-01-2018 | -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. Média Móvel

| data | inscrições |
| ------------ | ---------- |
| 01-01-2018 | 10
| 02-01-2018 | 20
| 03-01-2018 | 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. Várias condições de conexão

| id | assunto | de | para | timestamp |
| ---- | ---------- | -------------- | -------------- | --- ------------------ |
| 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 Correndo | 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 

Tarefas para funções da janela

Não. 1. Encontre o identificador com o valor máximo

  depname | empno | salário |     
----------- + ------- + -------- +
 desenvolver | 11 5200
 desenvolver | 7 4200
 desenvolver | 9 4500
 desenvolver | 8 6000
 desenvolver | 10 5200
 pessoal | 5 3500
 pessoal | 2 3900
 vendas | 3 4800
 vendas | 1 | 5000
 vendas | 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. Valor médio e classificação com uma função de janela (várias partes)

Parte 1

  depname | empno | salário |     
----------- + ------- + -------- +
 desenvolver | 11 5200
 desenvolver | 7 4200
 desenvolver | 9 4500
 desenvolver | 8 6000
 desenvolver | 10 5200
 pessoal | 5 3500
 pessoal | 2 3900
 vendas | 3 4800
 vendas | 1 | 5000
 vendas | 4 4800

  depname | empno | salário | avg_salary |     
----------- + ------- + -------- + ------------ +
 desenvolver | 11 5200 5020
 desenvolver | 7 4200 5020
 desenvolver | 9 4500 5020
 desenvolver | 8 6000 5020
 desenvolver | 10 5200 5020
 pessoal | 5 3500 3700
 pessoal | 2 3900 3700
 vendas | 3 4800 4867
 vendas | 1 | 5000 4867
 vendas | 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

Parte 2

  depname | empno | salário | salário_ranco |     
----------- + ------- + -------- + ------------- +
 desenvolver | 11 5200 2
 desenvolver | 7 4200 5
 desenvolver | 9 4500 4
 desenvolver | 8 6000 1 |
 desenvolver | 10 5200 2
 pessoal | 5 3500 2
 pessoal | 2 3900 1 |
 vendas | 3 4800 2
 vendas | 1 | 5000 1 |
 vendas | 4 4800 2

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

Outras tarefas de dificuldade média e alta

No. 1. Histogramas

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

| balde | contar |
| --------- | ------- |
| 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. Conexão cruzada (várias partes)

Parte 1

| estado | 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 

Parte 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. Cálculos avançados

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

| classe | contar |
| ------- | ------- |
| 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