_RoBust Okay. Enough. No more vowel questions, please! Be a little creative maybe? :/
felixschlitter I thought that 4 challenges ago. and they keep coming.
fengli97 hhhha , just pay little time to get much points :)
Bucci This guy gets it.
abhilashboharapi hahaha
ealmeyda Why doesnt anyone ever think of the letter Z?
marinskiy Here is Oracle solution from my HackerrankPractice repository:
SELECT DISTINCT City FROM Station WHERE REGEXP_LIKE(City, '^[^AEIOU].*[^aeiou]$');
Feel free to ask if you have any questions :)
pradeepINX Oracle :
SELECT distinct CITY FROM STATION WHERE not((SUBSTR(city,1,1)) in ('A','E','I','O','U')) and not(SUBSTR(city,length(trim(city)),1)) in ('a','e','i','o','u') order by city;
dwayneuni what does the TRIM mean ? and is there a more logical way to do it without TRIM ?
shubton My version for Oracle
select distinct city from station where regexp_like(city, '^[^aeiouAEIOU].*[^aeiouAEIOU]$');
Balaji003 Can u please explain?
dhirajt10 very nice solution
flashstorm use the 'i' flag in REGEXP_LIKE to reduce either of the capital letters
deepshah27996 [deleted]swsessaji A better one with case insensitive format
select distinct(city) from station where regexp_like(city,'^[^aeiou].*[^aeiou]$','i');
achintya0210 select distinct city from station where regexp_like(city, '^[^aeiou].*[^aeiou]$','i') order by city;
the 'i' in the regexp_like means case insensitive so no need to write them twice.
h597679894 My version for MySql
SELECT DISTINCT CITY FROM STATION WHERE CITY REGEXP '^[^AEIUO].*[^AEIOU]$'
akashingole1 what does that * do??
joseandradeferr1 ( * ) Matches zero or more occurrences.
buncreator u can replace * with any number of character
medipellipavan plz explian * and .
bhatushar .represents any singular character (it can be a digit, alphabet or a special character).*represents zero or more occurance of the previous character. So, when we use.*, we're basically asking for any number of occurance (even zero times) of any character.phanikumarvuka thanks for the clear explaination. But what is specifying start with or end with in the expression?
ashutoshtmr95 select DISTINCT CITY FROM STATION WHERE CITY NOT RLIKE '^[a,e,i,o,u]|[a,e,i,o,u]$'
ashutoshtmr95 select DISTINCT CITY FROM STATION WHERE CITY NOT RLIKE '^[a,e,i,o,u]|[a,e,i,o,u]$'
monikasezwar | why this OR symbol?? this query actually worked..but how? in questions its written "do not start with vowels and do not end with vowels", its AND not OR
dharmansh why you have used ^ symbol inside square brackets. please explain. I am writing the query like select distinct city from station where city not regexp '^[aeiou].*[aeiou]$'. This is not working :/
wnityanand why you used ^ 2 times?
arunsaini0111 ^[^AEIUO] one '^' is for starts with, another '^' is for doing reverse of that.
phanikumarvuka what do you mean by reverse of that. Could you explain?
stkirill [AEIUO] means 'any of these characters'
[^AEIUO] means 'any charachters except these ones'
It's kinda 'NOT' operator inside square brackets.
jadrake75 I had a few variations on that - I suppose if I used UPPER(CITY) it would be less work since most cities start with upper case letter (maybe some dutch do not!) - I had trouble with the ending group (I thought I could get it done with just one grouping at first)
SELECT DISTINCT(CITY) FROM STATION WHERE LOWER(CITY) REGEXP '^[^aeiou].*[^aeiou]$';
pradeepINX [deleted]pradeepINX trim means we need to remove the unexpected spaces from the outer edges of a string value.
shubton [deleted]chitluripallav can you explain in detail?
alexteel "Query the list of CITY names from STATION that do not start with vowels and do not end with vowels."
Should be
Query the list of CITY names from STATION that do not start or end with vowels.
ketansingh11 In MySQL:
select distinct city from station where city regexp '^[^aeiou].*[^aeiou]$'
mithun9421 Could you explain this for me? [^aeiou].* --> this thing(^) would simply negates everything ri8...then whhy should i use this for displaying everything that should start with a,e,i,o,u?
Thanks in advance
andrei_solares It is not a negation, the ^ is a complement which is different
sada_varti hope this will help you understand more regular expression
islamz1214 [^aeiou] -> should not match any character aeiou
^[^aeiou] -> should not match first character with aeiou
The ^ in brackets negates.
bahalekiitu like mine ;)
adarsh1392 A simple soln for MS SQL server:
select distinct city from station where (left(city,1) not in ('a','e','i','o','u') and right(city,1) not in ('a','e','i','o','u'));
Snehaniranjana This one works,
select distinct city from station where lower(substr(city,1,1)) not in ('a','e','i','o','u') and lower(substr(city,-1,1)) not in ('a','e','i','o','u');
Hope it helps
bhaskar61jha can you define the terms substr city(1,1) city(-1,1) regexp
chaurasiyavikas1 why you use lower before (substr(city,1,1))
dvt32 MySQL:
SELECT DISTINCT CITY FROM STATION WHERE CITY REGEXP '^[^aeiou]' AND CITY REGEXP '[^aeiou]$';
MaryGirgis Thanks it's working well
enmar or...
SELECT DISTINCT CITY
FROM STATION
WHERE CITY REGEXP '^[^aeiou].*[^aeiou]$'
haotie select distinct(city) from station where lower(city) not regexp '^[aeiou]|[aeiou]$'
kartikmathpal SELECT DISTINCT CITY FROM STATION WHERE CITY REGEXP '^[^aeiou].*[^aeiou]$'
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