Mar 5, 2016 · 10 Words beginning and ending with vowels \b[aeiou](\w*[aeiou])?\b \b stands for word boundary (in this case, beginning and ending of word). The optional grouping ()? is there to match …

Nov 14, 2013 · Here x is the key and the lenght of the list returned by the regex is the count of each vowel in this string, note that regex will find any character you introduce into the "aeiou" string.

WHERE name REGEXP '^[aeiou].*[aeiou]$' ^ and $ anchor the match to the beginning and end of the value. In my test, this won't use an index on the name column, so it will need to perform a full scan, …

I'm supposed to write a function that takes a character (i.e. a string of length 1) and returns true if it is a vowel, false otherwise. I came up with two functions, but don't know which one is bet...

I'm using basic JavaScript to count the number of vowels in a string. The below code works but I would like to have it cleaned up a bit. Would using .includes() help at all considering it is a st...

Mar 12, 2018 · Select distinct city from station where city not regexp '^[aeiou].*[aeiou]$' EXPLANATION : Using regular expression and NOT condition. So basically it is searches for the city which start with …

Feb 17, 2021 · In this case, it is being given the regular expression (?i)[^aeiou], which matches all non-vowels, and replaces them with "", an empty string. Then it returns the length of the resulting string,. …

Jun 13, 2019 · SELECT DISTINCT CITY FROM STATION WHERE CITY REGEXP '^[^aeiou].*[^aeiou]$'; Explanation: DISTINCT is used to remove duplicates from the result set of a …

Mar 19, 2022 · To get the number of unique vowels, you can lowercase the entire string, then call set() on the lowercased string to get a set of unique letters. You can then check whether each vowel is in …

for c in word: if c.lower() in vowels: word = word.replace(c,"") return word Explanation: I just changed if c in vowels: to if c.lower() in vowels: .lower() convert a string to lowercase. So it converted each letter …