✏️ docs: fix typos and rearrange questions

docs/challenging-sql-problems/challenging-sql-problems.md

@@ -19,9 +19,9 @@ As long as you know your database features and query patterns, these are fairly
 1. [Temperature anomaly detection](problems/bronze/temperature-anomaly-detection.md)
 2. [Personalised customer emails](problems/bronze/personalised-customer-emails.md)
 3. [Suspicious login activity](problems/bronze/suspicious-login-activity.md)
-4. [UK bank holidays](problems/bronze/uk-bank-holidays.md)
-5. [Customer churn](problems/bronze/customer-churn.md)
-6. [Fibonacci sequence](problems/bronze/fibonacci-sequence.md)
+4. [Customer churn](problems/bronze/customer-churn.md)
+5. [Fibonacci sequence](problems/bronze/fibonacci-sequence.md)
+6. [UK bank holidays](problems/bronze/uk-bank-holidays.md)
 
 ### ⚪ Silver Tier
 

docs/challenging-sql-problems/problems/bronze/customer-churn--sample-input.sql

@@ -1,8 +1,8 @@
 ```sql
 with user_history(user_id, last_update, activity_history) as (
     values
-        (1, '2024-03-01', 81),
-        (2, '2024-03-01', 2688),
+        (1, '2024-03-01',    81),
+        (2, '2024-03-01',  2688),
         (3, '2024-03-01', 13144)
 )
 ```

docs/challenging-sql-problems/problems/bronze/customer-churn.md

@@ -105,7 +105,6 @@ We can apply the same logic to the other bits:
 └-------------- 2024-02-24 (active)
 ```
 
-└─
 Since there are no more bits on the left, we can assume that the user was inactive on those days.
 
 This user is therefore not a churned user since they were active at least once in the last 7 days.
@@ -114,7 +113,7 @@ This user is therefore not a churned user since they were active at least once i
 
 The `activity_history` for user 2 is `2688`, which in binary is `101010000000`.
 
-We can expand the binary representation to understand the activity again:
+We can expand the binary representation to understand the activity:
 
 ```
 1 0 1 0 1 0 0 0 0 0 0 0
@@ -133,13 +132,13 @@ We can expand the binary representation to understand the activity again:
 └------------------------ 2024-02-19 (active)
 ```
 
-User 2 _has_ churned user since they were not active in the last 7 days (between `2024-02-24` and `2024-03-01`) but were active at least once in the 7 days before that.
+User 2 _has_ churned since they were not active in the last 7 days (between `2024-02-24` and `2024-03-01`) but were active at least once in the 7 days before that.
 
 #### User 3
 
 The `activity_history` for user 3 is `13144`, which in binary is `11001101011000`.
 
-We can expand the binary representation to understand the activity again:
+Again, we can expand the binary representation to understand the activity:
 
 ```
 1 1 0 0 1 1 0 1 0 1 1 0 0 0

docs/challenging-sql-problems/problems/bronze/customer-churn.sql

@@ -1,15 +1,15 @@
 ```sql
 create table user_history (
-    user_id int primary key,
-    last_update date not null,
+    user_id          int primary key,
+    last_update      date not null,
     activity_history bigint not null,
 );
 insert into user_history
 values
-    (1, '2024-06-01', 1056256),
+    (1, '2024-06-01',   1056256),
     (2, '2024-06-01', 907289368),
     (3, '2024-06-01', 201335032),
-    (4, '2024-06-01', 9769312),
+    (4, '2024-06-01',   9769312),
     (5, '2024-06-01', 246247510),
     (6, '2024-06-01', 492660983)
 ;

docs/challenging-sql-problems/problems/bronze/fibonacci-sequence--sample-output.sql

@@ -1,15 +1,15 @@
 ```sql
 solution(n, f_n) as (
     values
-        (1, 1),
-        (2, 1),
-        (3, 2),
-        (4, 3),
-        (5, 5),
-        (6, 8),
-        (7, 13),
-        (8, 21),
-        (9, 34),
+        (1,  1),
+        (2,  1),
+        (3,  2),
+        (4,  3),
+        (5,  5),
+        (6,  8),
+        (7,  13),
+        (8,  21),
+        (9,  34),
         (10, 55)
 )
 ```

docs/challenging-sql-problems/problems/bronze/fibonacci-sequence.md

@@ -61,4 +61,4 @@ The solution can be found at:
 <!-- prettier-ignore -->
 >? TIP: **Hint 2**
 >
-> Use the columns `n`, `f_n`, and `f_m` to keep track of the current term, the current Fibonacci number, and the previous Fibonacci number.
+> Use the columns `n`, `f_n`, and `f_m` to keep track of the current term, the current Fibonacci number, and the previous Fibonacci number, respectively.

docs/challenging-sql-problems/problems/bronze/personalised-customer-emails.sql

@@ -1,9 +1,9 @@
 ```sql
 create table customers (
-    customer_id int primary key,
-    full_name varchar not null,
-    first_name varchar,
-    last_name varchar,
+    customer_id   int primary key,
+    full_name     varchar not null,
+    first_name    varchar,
+    last_name     varchar,
     email_address varchar
 );
 insert into customers
@@ -25,8 +25,8 @@ values
 
 create table customer_relationships (
     parent_customer_id int references customers(customer_id),
-    child_customer_id int references customers(customer_id),
-    relationship_type varchar not null,
+    child_customer_id  int references customers(customer_id),
+    relationship_type  varchar not null,
     primary key (parent_customer_id, child_customer_id)
 );
 insert into customer_relationships

docs/challenging-sql-problems/problems/bronze/suspicious-login-activity--sample-input.sql

@@ -1,14 +1,14 @@
 ```sql
 with events(event_id, user_id, event_datetime, event_type) as (
     values
-        (1,  1, '2024-01-01 03:00:00', 'login failed'),
-        (2,  1, '2024-01-01 03:01:00', 'login failed'),
-        (3,  1, '2024-01-01 03:02:00', 'login failed'),
-        (4,  1, '2024-01-01 03:03:00', 'login failed'),
-        (5,  1, '2024-01-01 03:04:00', 'login failed'),
-        (6,  1, '2024-01-01 03:05:00', 'login'),
-        (7,  2, '2024-01-01 10:00:00', 'login'),
-        (8,  2, '2024-01-01 15:00:00', 'logout'),
-        (9,  2, '2024-01-01 23:00:00', 'login failed')
+        (1, 1, '2024-01-01 03:00:00', 'login failed'),
+        (2, 1, '2024-01-01 03:01:00', 'login failed'),
+        (3, 1, '2024-01-01 03:02:00', 'login failed'),
+        (4, 1, '2024-01-01 03:03:00', 'login failed'),
+        (5, 1, '2024-01-01 03:04:00', 'login failed'),
+        (6, 1, '2024-01-01 03:05:00', 'login'),
+        (7, 2, '2024-01-01 10:00:00', 'login'),
+        (8, 2, '2024-01-01 15:00:00', 'logout'),
+        (9, 2, '2024-01-01 23:00:00', 'login failed')
 )
 ```

docs/challenging-sql-problems/problems/gold/loan-repayment-schedule.md

@@ -17,7 +17,7 @@
 > - `repayment_date` as the date of the repayment
 > - `interest` as the interest component of the repayment
 > - `principal` as the principal component of the repayment
-> - `total` as the total repayment
+> - `total` as the total value of the repayment
 > - `balance` as the outstanding balance _after_ the repayment
 >
 > Order the output by `loan_id` and `repayment_number`.

docs/challenging-sql-problems/problems/gold/travel-plans--sample-input.sql

@@ -14,7 +14,7 @@ routes_schedule(
 ) as (
     values
         (1, 'train', 'London Gatwick', 'London St Pancras', '09:00:00+00:00', '21:00:00+00:00', '01:00:00', '00:30:00', 12.25),
-        (2, 'bus',   'London Gatwick', 'London St Pancras', '06:00:00+00:00', '06:00:00+00:00', null,       '03:30:00', 8.50)
+        (2, 'bus',   'London Gatwick', 'London St Pancras', '06:00:00+00:00', '06:00:00+00:00', null,       '03:30:00',  8.50)
 ),
 
 routes_timetable(
@@ -29,6 +29,6 @@ routes_timetable(
     values
         (1, 'boat',  'New York',          'London Gatwick', '2024-01-01 04:30:00-05:00', '2024-01-01 22:00:00+00:00', 179.00),
         (2, 'plane', 'New York',          'London Gatwick', '2024-01-01 18:00:00-05:00', '2024-01-02 10:45:00+00:00', 125.00),
-        (3, 'boat',  'London St Pancras', 'Paris',          '2024-01-02 13:00:00+00:00', '2024-01-02 14:30:00+01:00', 75.00)
+        (3, 'boat',  'London St Pancras', 'Paris',          '2024-01-02 13:00:00+00:00', '2024-01-02 14:30:00+01:00',  75.00)
 )
 ```

docs/challenging-sql-problems/problems/gold/travel-plans.md

@@ -2,9 +2,9 @@
 
 > [!SUCCESS] Scenario
 >
-> You're helping a client plan a trip from New York to Paris, and they want you to find the fastest and cheapest route.
+> You're helping a client plan a trip from New York to Paris, and they want you to find the fastest and cheapest routes.
 >
-> They have collected journey information into two tables:
+> They have collected journey information for routes that they are happy to take into two tables:
 >
 > - `routes_timetable`
 > - `routes_schedule`
@@ -17,7 +17,7 @@
 >
 > Given the tables that your client has provided, find the fastest and cheapest route from New York to Paris, leaving after `2024-01-01 12:00:00-05:00`.
 >
-> Give a minimum of 30 minutes and a maximum of 2 hours for "interchange" time (the time between arrival and departure at the same location). All costs are in the same currency (with no currency specified by the client).
+> Give a minimum of 30 minutes and a maximum of 6 hours for "interchange" time, which is the time between arrival and departure at the same location. All costs are in the same currency (with no currency specified by the client).
 >
 > The output should have at most two rows (the fastest/cheapest routes, which may be the same route), with the columns:
 >
@@ -50,7 +50,7 @@ The solution can be found at:
 > |           1 | train             | London Gatwick | London St Pancras | 09:00:00 +00:00    | 21:00:00 +00:00  | 01:00:00   | 00:30:00 | 12.25 |
 > |           2 | bus               | London Gatwick | London St Pancras | 06:00:00 +00:00    | 06:00:00 +00:00  | _null_     | 03:30:00 |  8.50 |
 >
-> **Routes Schedule**
+> **Routes Timetable**
 >
 > | route_id | mode_of_transport | from_location     | to_location    | departure_datetime | arrival_datetime  |   cost |
 > |---------:|:------------------|:------------------|:---------------|:-------------------|:------------------|-------:|

docs/challenging-sql-problems/problems/gold/travel-plans.sql

@@ -1,47 +1,47 @@
 ```sql
 create table routes_schedule (
-    schedule_id int primary key,
-    mode_of_transport varchar not null,
-    from_location varchar not null,
-    to_location varchar not null,
+    schedule_id        int primary key,
+    mode_of_transport  varchar not null,
+    from_location      varchar not null,
+    to_location        varchar not null,
     earliest_departure timetz not null,
-    latest_departure timetz not null,
-    frequency time,  /* `null` means that it's daily */
-    duration time not null,
-    cost decimal(8, 2) not null,
+    latest_departure   timetz not null,
+    frequency          time,  /* `null` means that it's daily */
+    duration           time not null,
+    cost               decimal(8, 2) not null,
 );
 insert into routes_schedule
 values
     (1, 'train', 'London St Pancras', 'London Gatwick',    '08:00:00+00:00', '20:00:00+00:00', '01:00:00', '00:30:00', 17.50),
     (2, 'train', 'London St Pancras', 'London Gatwick',    '07:30:00+00:00', '22:30:00+00:00', '02:30:00', '01:15:00', 12.00),
-    (3, 'bus',   'London St Pancras', 'London Gatwick',    '06:15:00+00:00', '06:15:00+00:00', null,       '03:30:00', 6.75),
-    (4, 'bus',   'London St Pancras', 'London Gatwick',    '19:30:00+00:00', '19:30:00+00:00', null,       '03:30:00', 6.75),
+    (3, 'bus',   'London St Pancras', 'London Gatwick',    '06:15:00+00:00', '06:15:00+00:00', null,       '03:30:00',  6.75),
+    (4, 'bus',   'London St Pancras', 'London Gatwick',    '19:30:00+00:00', '19:30:00+00:00', null,       '03:30:00',  6.75),
     (5, 'train', 'London Gatwick',    'London St Pancras', '09:00:00+00:00', '21:00:00+00:00', '01:00:00', '00:30:00', 17.50),
     (6, 'train', 'London Gatwick',    'London St Pancras', '07:15:00+00:00', '22:15:00+00:00', '02:30:00', '01:15:00', 12.00),
-    (7, 'bus',   'London Gatwick',    'London St Pancras', '06:00:00+00:00', '06:00:00+00:00', null,       '03:30:00', 6.75),
-    (8, 'bus',   'London Gatwick',    'London St Pancras', '20:00:00+00:00', '20:00:00+00:00', null,       '03:30:00', 6.75)
+    (7, 'bus',   'London Gatwick',    'London St Pancras', '06:00:00+00:00', '06:00:00+00:00', null,       '03:30:00',  6.75),
+    (8, 'bus',   'London Gatwick',    'London St Pancras', '20:00:00+00:00', '20:00:00+00:00', null,       '03:30:00',  6.75)
 ;
 
 create table routes_timetable (
-    route_id int primary key,
-    mode_of_transport varchar not null,
-    from_location varchar not null,
-    to_location varchar not null,
+    route_id           int primary key,
+    mode_of_transport  varchar not null,
+    from_location      varchar not null,
+    to_location        varchar not null,
     departure_datetime timestamptz not null,
-    arrival_datetime timestamptz not null,
-    cost decimal(8, 2) not null,
+    arrival_datetime   timestamptz not null,
+    cost               decimal(8, 2) not null,
 );
 insert into routes_timetable
 values
-    (1,  'boat',  'London St Pancras', 'Paris',          '2024-01-01 06:00:00+00:00', '2024-01-01 07:30:00+01:00', 45.00),
+    (1,  'boat',  'London St Pancras', 'Paris',          '2024-01-01 06:00:00+00:00', '2024-01-01 07:30:00+01:00',  45.00),
     (2,  'plane', 'London Gatwick',    'New York',       '2024-01-01 13:05:00+00:00', '2024-01-01 20:55:00-05:00', 158.00),
     (3,  'plane', 'London Gatwick',    'New York',       '2024-01-02 20:40:00+00:00', '2024-01-03 04:30:00-05:00', 147.00),
-    (4,  'plane', 'London St Pancras', 'Paris',          '2024-01-03 07:00:00+00:00', '2024-01-03 08:30:00+01:00', 70.00),
+    (4,  'plane', 'London St Pancras', 'Paris',          '2024-01-03 07:00:00+00:00', '2024-01-03 08:30:00+01:00',  70.00),
     (5,  'plane', 'Paris',             'New York',       '2024-01-02 12:00:00+01:00', '2024-01-02 20:30:00-05:00', 180.00),
     (6,  'plane', 'New York',          'London Gatwick', '2024-01-01 13:00:00-05:00', '2024-01-02 05:45:00+00:00', 160.00),
     (7,  'boat',  'New York',          'London Gatwick', '2024-01-01 05:30:00-05:00', '2024-01-01 23:00:00+00:00', 195.00),
-    (8,  'boat',  'London St Pancras', 'Paris',          '2024-01-01 18:00:00+00:00', '2024-01-01 19:30:00+01:00', 95.00),
-    (9,  'boat',  'London St Pancras', 'Paris',          '2024-01-02 14:00:00+00:00', '2024-01-02 15:30:00+01:00', 40.00),
+    (8,  'boat',  'London St Pancras', 'Paris',          '2024-01-01 18:00:00+00:00', '2024-01-01 19:30:00+01:00',  95.00),
+    (9,  'boat',  'London St Pancras', 'Paris',          '2024-01-02 14:00:00+00:00', '2024-01-02 15:30:00+01:00',  40.00),
     (10, 'plane', 'New York',          'Paris',          '2024-01-01 18:00:00-05:00', '2024-01-02 17:45:00+01:00', 279.00)
 ;
 ```

docs/challenging-sql-problems/problems/silver/bus-routes.sql

@@ -1,8 +1,8 @@
 ```sql
 create table bus_stops (
-    bus_id int,
+    bus_id    int,
     from_stop varchar,
-    to_stop varchar,
+    to_stop   varchar,
     primary key (bus_id, from_stop)
 );
 insert into bus_stops

docs/challenging-sql-problems/problems/silver/decoding-datelist-ints.md

@@ -10,6 +10,8 @@
 >
 > The number of rows to show in the output should be the number of days in the `activity_history` column for the user with the most days.
 >
+> Just like in the [customer churn](../bronze/customer-churn.md) problem, the `last_update` column will always have the same date for all users.
+>
 > The output should have a single row per day with the columns:
 >
 > - `active_date` as the date of the activity, starting from the `last_update` date

docs/challenging-sql-problems/problems/silver/funnel-analytics--sample-output.sql

@@ -1,13 +1,13 @@
 ```sql
 solution(cohort, stage, mortgages, step_rate, total_rate) as (
     values
-        ('2024-01', 'full application',     1, 100, 100),
-        ('2024-01', 'decision',             1, 100, 100),
-        ('2024-01', 'documentation',        1, 100, 100),
-        ('2024-01', 'valuation inspection', 1, 100, 100),
-        ('2024-01', 'valuation made',       1, 100, 100),
-        ('2024-01', 'valuation submitted',  1, 100, 100),
-        ('2024-01', 'solicitation',         1, 100, 100),
-        ('2024-01', 'funds released',       0,   0,   0)
+        ('2024-01', 'full application',     1, 100.00, 100.00),
+        ('2024-01', 'decision',             1, 100.00, 100.00),
+        ('2024-01', 'documentation',        1, 100.00, 100.00),
+        ('2024-01', 'valuation inspection', 1, 100.00, 100.00),
+        ('2024-01', 'valuation made',       1, 100.00, 100.00),
+        ('2024-01', 'valuation submitted',  1, 100.00, 100.00),
+        ('2024-01', 'solicitation',         1, 100.00, 100.00),
+        ('2024-01', 'funds released',       0,   0.00,   0.00)
 )
 ```

docs/challenging-sql-problems/problems/silver/funnel-analytics.md

@@ -29,6 +29,8 @@
 > - `step_rate` as the percentage of mortgages that reached the stage compared to the previous stage
 > - `total_rate` as the percentage of mortgages that reached the stage compared to the first stage
 >
+> Round the `step_rate` and `total_rate` to two decimal places.
+>
 > Note that each cohort should have _all_ the stages, even if there are no mortgages that reached that stage -- the `mortgages` column should be `0` in that case.
 >
 > Order the output by `cohort` and _the `stage` order_ (e.g. `full application` should come before `decision`, and so on).
@@ -64,14 +66,14 @@ The solution can be found at:
 >
 > | cohort  | stage                | mortgages | step_rate | total_rate |
 > |:--------|:---------------------|----------:|----------:|-----------:|
-> | 2024-01 | full application     |         1 |       100 |        100 |
-> | 2024-01 | decision             |         1 |       100 |        100 |
-> | 2024-01 | documentation        |         1 |       100 |        100 |
-> | 2024-01 | valuation inspection |         1 |       100 |        100 |
-> | 2024-01 | valuation made       |         1 |       100 |        100 |
-> | 2024-01 | valuation submitted  |         1 |       100 |        100 |
-> | 2024-01 | solicitation         |         1 |       100 |        100 |
-> | 2024-01 | funds released       |         0 |         0 |          0 |
+> | 2024-01 | full application     |         1 |    100.00 |     100.00 |
+> | 2024-01 | decision             |         1 |    100.00 |     100.00 |
+> | 2024-01 | documentation        |         1 |    100.00 |     100.00 |
+> | 2024-01 | valuation inspection |         1 |    100.00 |     100.00 |
+> | 2024-01 | valuation made       |         1 |    100.00 |     100.00 |
+> | 2024-01 | valuation submitted  |         1 |    100.00 |     100.00 |
+> | 2024-01 | solicitation         |         1 |    100.00 |     100.00 |
+> | 2024-01 | funds released       |         0 |      0.00 |       0.00 |
 >
 --8<-- "docs/challenging-sql-problems/problems/silver/funnel-analytics--sample-output.sql"
 

docs/challenging-sql-problems/problems/silver/mandelbrot-set.md

@@ -6,7 +6,7 @@
 
 > [!QUESTION]
 >
-> Plot an image of the Mandelbrot set.
+> Plot an image of the [Mandelbrot set](https://en.wikipedia.org/wiki/Mandelbrot_set).
 >
 > Start with a 51x51 grid of points ranging from -2 to 2 on both the x and y axes. A sample of some points in the grid are:
 >
@@ -51,7 +51,7 @@ The solution can be found at:
 
 - [mandelbrot-set.md](../../solutions/silver/mandelbrot-set.md)
 
-A worked example is provided below to help illustrate the loan calculations.
+A worked example is provided below to help illustrate the Mandelbrot set calculations.
 
 ---
 

docs/challenging-sql-problems/problems/silver/predicting-values--sample-input.sql

@@ -1,7 +1,7 @@
 ```sql
 with datasets(dataset_1__x, dataset_1__y, dataset_2__x, dataset_2__y) as (
     values
-        (1, 2.00, 1, 9.12),
+        (1, 2.00, 1,  9.12),
         (2, 4.00, 3, 31.18),
         (3, 6.00, 5, 55.27),
         (4, 8.00, 7, 61.12)