rshaghoulian MySQL solution
From my HackerRank solutions.
SELECT DISTINCT CITY FROM STATION WHERE CITY REGEXP '^[aeiou].*[aeiou]$';
Let me know if you have any questions.
JackySatpal why can't we do like the below? CITY REGEXP '^[aeiou]$'
rshaghoulian [deleted]Ippo343 Nope, it's because the regex is wrong.
^ // string's beginning [aeiou] // a SINGLE vowel $ // end of string
Therefore this would match the vowels and only the vowels. Even a string like "aaa" would not match, because it's more than 1 char long.
The correct one is the one you used at the beginning because of the ".*" bit:
^ // start of string [aeiou] // a single vowel . // any characted... * // ...repeated any number of times [aeiou] // another vowel $ // end of string
fjhahn This was helpful... Thanks for the in depth explanation
youssefaabdo Thanks
ssrini21 Why does not below condition work?
regexp_like (column_name, '^[aeiou]*[aeiou]$','i')
TurboPascal Wow, thank you :)
TomYumNoodle hello,
So why doesn't this work?
select distinct city from station where city regrex '^[aeiou]|[aeiou]$';
I followed the example used here: https://www.tutorialspoint.com/mysql/mysql-regexps.htm
Thank you!
psatishpatro88 if u write ''^[aeiou]|[aeiou]$', it will check only start letter vowel or end letter vowel like abcdfsd, efdsf, osdf(staring one vowel), dsfa, dsde, fghfgo(ending one vowel)
if u remove the pipeline, '^[aeiou][aeiou]$' it will check only start letter vowel end letter vowel and nothing between them
if u write '^[aeiou].[aeiou]$'; adding one dot means start letter vowel or end letter vowel and any one char between them
if u write '^[aeiou].*[aeiou]$'; adding one * means start letter vowel or end letter vowel and any one character repeated 0 or more than 0
Hope, u get it
TomYumNoodle Thank you so much!
psatishpatro88 You are welcome
gaurisankarj1 You're asked for the one's which start and end in vowels, you've only accounted for the one's starting, it would either be an error or the $ will not function like you'd like it to.
michelle__K can we use + instead of *??
rshaghoulian - passes the testcase but it's technically not correct since it would match strings of a longer length. * is 0 or more of a character. + is 1 or more of a character. As long as you know the difference between the 2 symbols you should be fine.
vyomeshaudi [deleted]vyomeshaudi [deleted]vyomeshaudi can you plz tell what does .* is used to do in your code
rshaghoulian Hi. According to this reference, The "." matches "any single character". The "*", matches "Zero or more instances of preceding element". Together, they match 0 or more of any character.
divine_soul It works without .* also.Is it right?
rshaghoulian I tried it without the .* as you suggested and it does not work.
divine_soul SELECT DISTINCT city FROM station WHERE city REGEXP '^[aeiou]' AND city REGEXP '[aeiou]$';
Actually i did it this way,i forgot to mention.
rshaghoulian Nice solution
skarthikamca Why my query is not working ?
SELECT DISTINCT CITY FROM STATION WHERE CITY LIKE 'a%a' OR CITY LIKE 'e%e' OR CITY LIKE 'i%i' OR CITY LIKE 'o%o' OR CITY LIKE 'u%u'
rshaghoulian Because a city name can start with "a" and end with "e", which your code does not account for. The first and last letters do not necessarily have to be the same.
dhrubodeepbasum1 Because you did not check all the permutation of vowels, there are more cases like aa,ae,ai,ao....etc.
isolsgrouptesti1 its not working i was implement same logic
mprksh1047 Can you please explain this piece of code in little more detail? Thanks
rshaghoulian https://www.hackerrank.com/challenges/weather-observation-station-8/forum/comments/399775 - this user comment explains it better than I could explain it.
nitishkumar8195 SELECT DISTINCT CITY FROM STATION WHERE CITY REGEXP '[aeiou]$.*^[aeiou]'; what is wrong in this ?
rex_liu_y select distinct city from station where left(city,1) in ('a','e','i','o','u') and right(city, 1) in ('a','e','i','o','u')
ajayverma1244 SMART
deepsammanna select DISTINCT CITY from station where city regexp '[aeiou]$' and city in (SELECT CITY FROM STATION WHERE CITY REGEXP '^[aeiou]');
deepsammanna will this have a performance issue?
evil_eyeguy simplified by taking out the nested SELECT and just replacing it with another REGEXP.
So we have for mySQL:
SELECT DISTINCT city FROM station WHERE city RLIKE '^[aeiou]' AND city RLIKE '[aeiou]$'
RLIKE is just a synonym for REGEXP for those who don't know.
Also, simplified even more by (credit: rshaghoulian):
SELECT DISTINCT city FROM station WHERE city RLIKE '^[aeiou.*aeiou]$'
rohit167822 this doesn't work for oracle, is it for some other database server? If yes, then of which?
phaberest I wrote the same but it doesn't work. Weird, I just ran it in my local database on a real table and there it just works.
It is not as clean as the regex version mentioned in the discussion, but well...
hacknik246 This is in Oracle SQL without REGEX using only basic SELECT statements. It's easy to read. Hope it helps.
SELECT city FROM station WHERE ( LOWER(city) LIKE 'a%' OR LOWER(city) LIKE 'e%' OR LOWER(city) LIKE 'i%' OR LOWER(city) LIKE 'o%' OR LOWER(city) LIKE 'u%' ) AND ( LOWER(city) LIKE '%a' OR LOWER(city) like '%e' OR LOWER(city) LIKE '%i' OR LOWER(city) LIKE '%o' OR LOWER(city) LIKE '%u' );
Tmtorres82 [deleted]
stacyjay2008 select DISTINCT (city) from STATION where REGEXP_LIKE(city, '^[aeiou]','i') and REGEXP_LIKE(city, '[aeiou]$','i');
using regular expression is easier
avinasp [deleted]avinasp @hacknik246 Terrfic, after many trials, your code worked! Thanks!
marinskiy Here is Oracle solution from my HackerrankPractice repository:
SELECT DISTINCT CITY FROM STATION WHERE REGEXP_LIKE(City, '^[AEIOU].*[aeiou]$');
Feel free to ask if you have any questions :)
Carlosdlr check this one works with the oracle compiler version using by this site
SELECT DISTINCT(city) FROM station WHERE REGEXP_LIKE (lower(city), '^[aeiou]') and REGEXP_LIKE (lower(city), '[aeiou]$');
ICPhi This can be simplied I believe but here is my =soultion for MySQL Luckily left doesnt get greedy in these type of situations.
select distinct city from station where right(city, 1) in ('a','e','i','o','u') and left(city, 1) in ('a','e','i','o','u');
k_sharnitha It's Working.. SELECT DISTINCT CITY FROM STATION WHERE LOWER(SUBSTR(CITY,1,1)) in ('a','e','i','o','u') and LOWER(SUBSTR(CITY,LENGTH(CITY),1)) in ('a','e','i','o','u');
sailakshmi540 This is also working... thk u fr ur code ...this helped me a lott..
select distinct CITY from STATION where substr(UPPER(CITY),1,1) in ('A','E','I','O','U') and substr(UPPER(CITY),-1,1) in ('A','E','I','O','U');
gokhansavas90 SELECT distinct City FROM STATION WHERE City LIKE '[A,E,I,O,U]%[A,E,I,O,U]'
AKSHAY97 Not working in mysql
rohit167822 Not working in ORACLE
adia_alderson SELECT DISTINCT STATION.CITY FROM STATION WHERE STATION.CITY REGEXP '^[aeiouAEIOU].*[aeiouAEIOU]$'
vyomeshaudi can you plz tell what does .* is used to do in your code
adia_alderson Sure no problem. So .* is a greedy catch all for a regular expression. So basically I'm saying the first letter must be a vowel ^[aeiouAEIOU] then I don't care whats in between .* and then the last letter must be a vowel [aeiouAEIOU]$
vyomeshaudi ^[aeiouAEIOU]$ why cant i write it like this?
adia_alderson So the symbol ^ means the beginning of the match and the $ symbol means the end of the match. So when you do ^[aeiouAEIOU]$ This will match: A <--- match Apple <--- No match because your only allowing one character vowel between the beginning ^ and end $ of the match.
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