Query rows where first_name contains at least 2 vowels, and the number of occurences of each vowel is equal - sql

Question

I have the following problem: Show all rows in table where column first_name contains at least 2 vowels (a, e, i, o, u), and the number of occurences of each vowel is the same.
Valid example: Alexander, "e" appears 2 times, "a" appears 2 times. That is coreect.
Invalid example: Jonathan, it has 2 vowels (a, o), but "o" appears once, and "a" appears twice, the number of occurences is not equal.
I've solved this problem by calculating each vowel, and then verify every case (A E, A I, A O etc. Shortly, each combination of 2, 3, 4, 5). With that solution, I have a very long WHERE. Is there any shorter way and more elegant and simple?

Answer 1

This is how I solved it in TSQL in MS SQL Server 2019.
I know its not exactly what you wanted. Just an interesting thing to try. Thanks for that.
DROP TABLE IF EXISTS #Samples
SELECT n.Name
INTO #Samples
FROM
(
SELECT 'Ravi' AS Name
UNION
SELECT 'Tim'
UNION
SELECT 'Timothe'
UNION
SELECT 'Ian'
UNION
SELECT 'Lijoo'
UNION
SELECT 'John'
UNION
SELECT 'Jami'
) AS n
SELECT g.Name,
IIF(MAX (g.Repeat) = MIN (g.Repeat) AND SUM (g.Appearance) >= 2, 'Valid', 'Invalid') AS Validity
FROM
(
SELECT v.value,
s.Name,
SUM (LEN (s.Name) - LEN (REPLACE (s.Name, v.value, ''))) AS Repeat,
SUM (IIF(s.Name LIKE '%' + v.value + '%', 1, 0)) AS Appearance
FROM STRING_SPLIT('a,e,i,o,u', ',') AS v
CROSS APPLY #Samples AS s
GROUP BY v.value,
s.Name
) AS g
WHERE g.Repeat > 0
GROUP BY g.Name
Output
we can replace STRING_SPLIT with a temp table for supporting lower versions
DROP TABLE IF EXISTS #Vowels
SELECT C.Vowel
INTO #Vowels
FROM
(
SELECT 'a' AS Vowel
UNION
SELECT 'e'
UNION
SELECT 'i'
UNION
SELECT 'o'
UNION
SELECT 'u'
) AS C
SELECT g.Name,
IIF(MAX (g.Repeat) = MIN (g.Repeat) AND SUM (g.Appearance) >= 2, 'Valid', 'Invalid') AS Validity
FROM
(
SELECT v.Vowel,
s.Name,
SUM (LEN (s.Name) - LEN (REPLACE (s.Name, v.Vowel, ''))) AS Repeat,
SUM (IIF(s.Name LIKE '%' + v.Vowel + '%', 1, 0)) AS Appearance
FROM #Vowels AS v
CROSS APPLY #Samples AS s
GROUP BY v.Vowel,
s.Name
) AS g
WHERE g.Repeat > 0
GROUP BY g.Name

Answer 2

From Oracle 12, you can use:
SELECT name
FROM table_name
CROSS JOIN LATERAL(
SELECT 1
FROM (
-- Step 2: Count the frequency of each vowel
SELECT letter,
COUNT(*) As frequency
FROM (
-- Step 1: Find all the vowels
SELECT REGEXP_SUBSTR(LOWER(name), '[aeiou]', 1, LEVEL) AS letter
FROM DUAL
CONNECT BY LEVEL <= REGEXP_COUNT(LOWER(name), '[aeiou]')
)
GROUP BY letter
)
-- Step 3: Filter out names where the number of vowels are
-- not equal or the vowels do not occur at least twice
-- and there are not at least 2 different vowels.
HAVING MIN(frequency) >= 2
AND MIN(frequency) = MAX(frequency)
AND COUNT(*) >= 2
);
Which, for the sample data:
CREATE TABLE table_name (name) AS
SELECT 'Alexander' FROM DUAL UNION ALL
SELECT 'Johnaton' FROM DUAL UNION ALL
SELECT 'Anna' FROM DUAL;
Outputs:
NAME
Alexander
db<>fiddle here