I was solving a problem on SQL from Hackerrank. I have to make a query such that it gives me all city names starting with a, e, i, o or u. I'm using Like operator but still wrong answer.
Here's the problem Link
Here's my solution-
SELECT DISTINCT CITY
FROM STATION
WHERE CITY LIKE '[AEIOU]%'
Can anybody explain?
LIKE does not support that parttern. You need a regular expression match for this:
SELECT DISTINCT CITY
FROM STATION
WHERE CITY REGEXP '^[AEIOU]'
In the regex, ^ represents the beginning of the string, and the square brackets define a custom character class (meaning that the first character must belong to that list).
On the other hand, if you were to use LIKE, you would need multiple conditions, which would make the code lengthier (and probably less efficient):
WHERE
CITY LIKE 'A%'
OR CITY LIKE 'E%'
OR CITY LIKE 'I%'
OR CITY LIKE 'O%'
OR CITY LIKE 'U%'
select distinct city from station where Left(city,1)='A' or Left(city,1)='E' or Left(city,1)='I' or Left(city,1)='O' or Left(city,1)='U';
This worked fine with me (MS SQL)
SELECT DISTINCT CITY FROM STATION WHERE LEFT(CITY, 1) IN ('A', 'E', 'I', 'O', 'U');
This return the City names starting with vowel.
Related
Query the list of CITY names starting with vowels (i.e., a, e, i, o, or u) from STATION. Your result cannot contain duplicates
select distinct city from station where city LIKE '[a,e,i,o,u]%'
You have two ways of doing this. If you want to use LIKE, you'll need to have separate checks for each starting vowel:
SELECT DISTINCT city
FROM station
WHERE city LIKE 'a%' OR city LIKE 'e%' OR city LIKE 'i%' OR
city LIKE 'u%' OR city LIKE 'u%';
If you can use REGEXP, it is possible to write the above logic more succinctly:
SELECT DISTINCT city
FROM station
WHERE city REGEXP '^[aeiou]'; -- ^ means starts with a vowel letter
This question already has answers here:
Query the list of CITY names from STATION that do not start with vowels and do not end with vowels
(22 answers)
Closed 6 months ago.
I'm doing exercises (SQL beginner practice) on hackerrank.com, which I'm sure many of you are familiar with.
This is the question:
Query the list of CITY names from STATION that do not start with vowels and do not end with vowels. Your result cannot contain duplicates.
My answer was the following:
SELECT DISTINCT(city)
FROM station
WHERE city NOT LIKE 'A%'
AND city NOT LIKE 'E%'
AND city NOT LIKE 'I%'
AND city NOT LIKE 'O%'
AND city NOT LIKE 'U%'
AND city NOT LIKE '%a'
AND city NOT LIKE '%e'
AND city NOT LIKE '%i'
AND city NOT LIKE '%o'
AND city NOT LIKE '%u'
I know this is a ridiculous way to go about it- can you guys offer suggestions on how to make this shorter? I'm just not knowledgable enough in SQL at the moment to make it more succinct. If you could show a couple ways to go about it, that would be really helpful.
In MySQL, you can use the REGEXP operator:
select distinct city
from station
where city not regexp '^[aeiou].*[aeiou]$'
If your column has a case-sensitive collation, then you can lower the value before comparison:
where lower(city) not regexp '^[aeiou].*[aeiou]$'
Or, if you are running MySQL 8.0, you can use regexp_like() and set the match argument to 'i' to make the search case-insensitive:
where not regexp_like(city, '^[aeiou].*[aeiou]$', 'i')
Something like this may help
SELECT DISTINCT city
FROM station
WHERE city NOT LIKE '[AEIOUaeiou]%[AEIOUaeiou]'
However, this answer is not %100 correct without knowing exact RDBMS system which you are using
If regex is an option then check that the city does not match the following pattern:
^[aeiou]|[aeiou]$
This pattern matches ax, xa, aa and a i.e. "city starts with or ends with a vowel". When you invert the result it becomes "city does not start with and does not end with a vowel". See De Morgan's law.
MySQL implementation:
SELECT DISTINCT city, NOT city REGEXP '^[aeiou]|[aeiou]$' AS result
FROM (
SELECT 'xx' AS city UNION
SELECT 'ax' UNION
SELECT 'xa' UNION
SELECT 'aa' UNION
SELECT 'x' UNION
SELECT 'a'
) AS station
-- returns 1 for xx and x, 0 for all others
#Dervis - This might be a bit faster.
SELECT DISTINCT(city)
FROM station
WHERE UPPER(city) NOT LIKE '[AEIOU]%[AEIOU]'
Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. Your result cannot contain duplicates.
THIS IS FOR MS SQL USERS
SELECT DISTINCT CITY FROM STATION WHERE CITY LIKE '[^aeiou]%' OR CITY LIKE '%[^aeiou]';
This is a challenge question from HackerRank.
My first query is working fine:
select distinct city
from station
where city not regexp '^[aeiou]'
and city not regexp '[aeiou]$'
But my second query is giving wrong answer
select distinct city
from station
where city not regexp '^[aeiouAEIOU].*[aeiouAEIOU]$'
It is compiling but giving wrong results
https://dev.mysql.com/doc/refman/8.0/en/regexp.html
'NOT REGEXP' is Negation of REGEXP,
'REGEXP' -> Checks the 'REGEXP' if satisfied then return 1 else 0
select 'Chelsea' REGEXP '^[aeiou].*[aeiou]$' ;
0
'NOT REGEXP' -> checks the 'REGEXP' then return the opposite of the result.
select 'Chelsea' not REGEXP '^[aeiou].*[aeiou]$' ;
1
In this case, 'Chelsea' doesn't start with vowel hence REGEXP is 0 and not REGEXP is 1.
Hence you are counting the cities, even the name starts or ends with vowels.
This works in MS SQL server :
SELECT DISTINCT city FROM station WHERE city NOT LIKE '[AEIOU]%' AND city NOT LIKE '%[aeiou]';
Try this it worked for me
select distinct city
from station
where city RLIKE'^[^aeiouAEIOU].*$';
SELECT DISTINCT city
FROM station
WHERE city NOT REGEXP '^[aeiouAEIOU].*|[aeiouAEIOU]$'
Explanation: I make sure the city does not start with ('^') vowels and also does not end with ('$') vowels.
In Mysql,
where not CITY Regexp '^[aeiou].*$' or not CITY Regexp '^.*[aeiou]$'
This worked for me using the RLIKE operator... Hope this helps:
select distinct CITY from STATION where CITY NOT RLIKE '^[aeiouAEIOU]' AND CITY NOT RLIKE '[AEIOUaeiou]$' GROUP BY CITY;
You need to take care of 3 cases when using any DB
1. Word starts with vowel but ends with consonant
2. Word starts with consonant but ends with vowel
3. Word starts with consonant and ends with consonant
my solution implemented in mySQL
select distinct(CITY) from station where CITY RLIKE '(^[^aeiouAEIOU].*[aeiou]$|^[aeiouAEIOU].*[^aeiou]$|^[^aeiouAEIOU].*[^aeiou]$)';
Use this MySQL Command:
SELECT DISTINCT CITY FROM STATION WHERE CITY REGEXP '^[^aeiou].*[^aeiou]$';
Explanation:
DISTINCT is used to remove duplicates from the result set of a SELECT statement.
"^[^aeiou]" means city names that do not start with vowels.
"[^aeiou]$" means city names that do not end with vowels.
".*" represents any string of any length.
Here is the right answer:
select DISTINCT(CITY)
from STATION
Where CITY RLIKE '^[^aieouAEIOU].*[^AEIOUaeiou]$';
Try this
select distinct city
from station
where city rlike '^[^aeiouAEIOU].*' and city rlike '.*[^aeiouAEIOU]$'
I consider both the answer right.
But some how first query is failed whereas second one is passed from hackerrank platform.
First query
SELECT DISTINCT city FROM station
WHERE city not RLIKE '^[aeiou].*[aeiou]$';
Second query : working with no issue
SELECT DISTINCT city FROM station
WHERE city not RLIKE '^[aeiou]'
and city not rlike '[aeiou]$';
SELECT DISTINCT city
FROM station
WHERE NOT city LIKE 'A%'
AND NOT city LIKE 'E%'
AND NOT city LIKE 'I%'
AND NOT city LIKE 'O%'
AND NOT city LIKE 'U%';
a lengthy way but you will get the answer
Here is a query using a Where column Not Rlike clause.
This does a pattern match of a string expression against a pattern.
Select Distinct city
From station
Where city Not Rlike '^[aeiouAEIOU].*[aeiouAEIOU]$';
Additionally, the Distinct is used to remove all results that are the same.
Query the list of CITY names from STATION that either do not start with vowels or do not end with vowels. Your result cannot contain duplicates.
THIS IS FOR MS SQL USERS
SELECT DISTINCT CITY FROM STATION WHERE CITY LIKE '[^aeiou]%' OR CITY LIKE '%[^aeiou]';
If regex is an option then check that the city does not match the following pattern:
^[aeiou]|[aeiou]$
This pattern matches ax, xa, aa and a i.e. "city starts with or ends with a vowel". When you invert the result it becomes "city does not start with and does not end with a vowel". See De Morgan's law.
MySQL implementation:
SELECT DISTINCT city, NOT city REGEXP '^[aeiou]|[aeiou]$' AS result
FROM (
SELECT 'xx' AS city UNION
SELECT 'ax' UNION
SELECT 'xa' UNION
SELECT 'aa' UNION
SELECT 'x' UNION
SELECT 'a'
) AS station
-- returns 1 for xx and x, 0 for all others
This can also work:
SELECT DISTINCT city
FROM station
WHERE city NOT RLIKE '[aeiou]$';
Simplest code, you can try:
select distinct city
from station
WHERE lower(city) Not Rlike '^[aeiou].*[aeiou]$';
OR
SELECT DISTINCT city FROM station
WHERE lower(city) not RLIKE '^[aeiou]'
and lower(city) not rlike '[aeiou]$';
This is basic solution for this problem, I hope it will surely work.
Select Distinct City From Station
Where City not LIKE 'a%a' and City not LIKE 'a%e' and City not LIKE 'a%i' and City not LIKE 'a%o' and City not LIKE 'a%u' and City not LIKE 'e%a' and City not LIKE 'e%e' and City not LIKE 'e%i' and City not LIKE 'e%o' and City not LIKE 'e%u' and City not LIKE 'i%a' and City not LIKE 'i%e' and City not LIKE 'i%i' and City not LIKE 'i%o' and City not LIKE 'i%u' and City not LIKE 'o%a' and City not LIKE 'o%e' and City not LIKE 'o%i' and City not LIKE 'o%o' and City not LIKE 'o%u' and City not LIKE 'u%a' and City not LIKE 'u%e' and City not LIKE 'u%i' and City not LIKE 'u%o' and City not LIKE 'u%u';
SELECT DISTINCT CITY FROM STATION WHERE NOT LOWER(LEFT(CITY, 1)) IN ('a', 'e', 'i', 'o', 'u') AND NOT LOWER(RIGHT(CITY, 1)) IN ('a', 'e', 'i', 'o', 'u');
--This works for db2 databases
I got the correct out put for the below code for MySql server
select distinct(city) from station
where substr(city,1,1) not in ('a','e','i','o','u') or substr(city,-1,1) not in ('a','e','i','o','u')
Most of us are reading the question wrong. They want either of the one condition to be fulfilled, not the both. The following solution works well in MS SQL Server :
select distinct city from station where (left(city,1) not in ('a','e','i','o','u') or right(city,1) not in ('a','e','i','o','u') );
select distinct city as cty from station where city not in (
select city from Station where
Right(city,1) in ('a','e','i','o','u') and
Left(city,1) in ('a','e','i','o','u')
)
Why is it not working?
Query the list of CITY names ending with vowels (a, e, i, o, u) from STATION. Our result cannot contain duplicates.
SELECT A DISTINCT CITY FROM THE STATION
WHERE CITY LIKE '%A' OR CITY LIKE '%E' OR LIKE '%I' LIKE '%O' OR LIKE '%U'
ORDER BY CITY;
What would be the correct query?
SELECT distinct CITY FROM STATION WHERE CITY LIKE '%a' OR CITY LIKE '%e' OR CITY LIKE '%i' OR CITY LIKE '%o' OR CITY LIKE '%u';
The correct answer is above 👆
this is accepted, as inside the quotations, the alphabet is considered case sensitive
hence use '%a' and similarly all the other vowels.
(% is a wildcard entry)
You can try this also-
SELECT DISTINCT CITY FROM STATION WHERE CITY REGEXP '[aeiouAEIOU]$';
I solved that problem this way. Instead of checking the last character one by one, I thought it is easier to check if we use in operator.
select distinct city
from station
where substring(city,len(city),1) in ('a','e','u','o','i')
order by city;
Using MySQL;
select distinct CITY
from STATION
where right(city,1) in ('a', 'e', 'i', 'o', 'u');
You query has syntax error, for each "or", you need to put city like instead of or. In addition, remember that in most of the dbms, it is case sensitive. If you want to ignore both upper case and lower case, do it like this. I am using mysql syntax, different dbms has different functions for lcase
SELECT DISTINCT CITY FROM STATION
WHERE lcase(CITY) LIKE '%a'
OR lcase(CITY) LIKE '%e'
OR lcase(CITY) LIKE '%i'
OR lcase(CITY) LIKE '%o'
OR lcase(CITY) LIKE '%u'
ORDER BY CITY;
Try with MySQL solution :
select distinct CITY from STATION where substr(CITY, -1, 1) in ('a','e','i','o','u');
Here "distinct" will solve the problem of duplicate value and "substring" function extract substring from string . Substring also contain start & length .
For more details follow the link :- https://www.w3schools.com/sql/func_mysql_substr.asp
I solved it like this:
SELECT DISTINCT CITY FROM STATION WHERE REGEXP_LIKE(CITY,'[aeiou]$','i');
In this case, the 'i' at the end is called a "flag" which means the vowels can be upper or lower case.
select distinct CITY from STATION where CITY like '%[a,e,i,o,u]';
Try this in MS SQL server
select distinct(city) from station where city like '%[a,e,i,o,u]' order by city;
select CITY
from STATION
where right(lower(CITY),1) in ('a','e','i','o','u')
group by CITY;
Using Regular Expression in MySQL would be like this:
SELECT DISTINCT(CITY) FROM STATION WHERE REGEXP_LIKE(CITY, '[aiueo]$', 'i') ORDER BY CITY;
you can use substring to solve that
SELECT DISTINCT CITY FROM STATION WHERE SUBSTRING(CITY,-1,1) NOT IN ('a','e','i','o','u')
For Mysql, this worked for me -
select DISTINCT(CITY) from STATION
where CITY REGEXP '[aeiou]$';
Select distinct city
from Station
where lower(SUBSTR(city,length(city),1))in ('a','e','i','o','u');
Simple but easy
select distinct(CITY)
from Station
where substr(City,LENGTH(CITY),Length(City)) in ('a','e','i','o','u')
In MsSql
Try this:
SELECT DISTINCT CITY FROM STATION WHERE RIGHT(CITY,1) IN ('a', 'e', 'i', 'o', 'u');
I want to print all the cities that do not start with a vowel, and they also must not end on a vowel.
I am aware that there is a lot of tutorial about how to do it when it must not start with a specific letter. But I am not able to find it, expressing several letters.
This is what I tried:
select distinct city
from station
where city not like '%[aeiou]%';
So if I say
where city not like '%a'
I will get the cities that do not end with the letter a.
Doing:
where city not like 'a%'
I will get the cities that do not start with the letter a
So my attempt with: where city not like '%[aeiou]%'; is to combine the two. Id does not throw an error, it does give me a list, but just not the correct result. Actually the list is so big, that I don't know what the expression I wrote above gives me. I tried several other attempts, but this is probably the most qualified one.
How can I get a list of all cities that do not start with, or end with a vowel?
You can simply use the SUBSTRING() function:
WHERE LOWER(SUBSTRING(city, 1, 1)) NOT IN ('a', 'e', 'i', 'o', 'u')
AND LOWER(SUBSTRING(city, -1, 1)) NOT IN ('a', 'e', 'i', 'o', 'u')
I also use the LOWER() function here, so that you don't have to repeat the same for upper case letters.
In MySQL, you would use regular expression. This pattern:
where city not like '%[aeiou]%';
will return all cities, because no city -- to my knowledge -- has a name with the sequence of characters '[aeiou]' in it. The above would do something more meaningful in SQL Server or Sybase, where those patterns are supported by like.
The regular expression looks like:
where city regexp '^[^aeiouAEIOU].*[^aeiouAEIOU]$'
I think this is what you're trying to do:
declare #Station table (city varchar(100))
insert into #station select 'Houston'
insert into #station select 'Astoria'
insert into #station select 'enigma'
select distinct
city
from
#Station
where
city not like '[aeiou]%[aeiou]'