Решить две задачи с codewars на Scala (2 любые из описания) 300 грн

Count the smiley faces!
Given an array (arr) as an argument complete the function countSmileys that should return the total number of smiling faces.

Rules for a smiling face:

• Each smiley face must contain a valid pair of eyes. Eyes can be marked as : or ;
• A smiley face can have a nose but it does not have to. Valid characters for a nose are - or ~
• Every smiling face must have a smiling mouth that should be marked with either ) or D

No additional characters are allowed except for those mentioned.

Valid smiley face examples: :) 😁 ;-D :~)
Invalid smiley faces: ;( :> :} :]

Example

countSmileys([':)', ';(', ';}', ':-D']);       // should return 2;

countSmileys([';D', ':-(', ':-)', ';~)']);     // should return 3;

countSmileys([';]', ':[', ';*', ':$', ';-D']); // should return 1;

Note

In case of an empty array return 0. You will not be tested with invalid input (input will always be an array). Order of the face (eyes, nose, mouth) elements will always be the same.

Count the number of Duplicates

Write a function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string. The input string can be assumed to contain only alphabets (both uppercase and lowercase) and numeric digits.

Example

"abcde" -> 0 # no characters repeats more than once
"aabbcde" -> 2 # 'a' and 'b'
"aabBcde" -> 2 # 'a' occurs twice and 'b' twice (`b` and `B`)
"indivisibility" -> 1 # 'i' occurs six times
"Indivisibilities" -> 2 # 'i' occurs seven times and 's' occurs twice
"aA11" -> 2 # 'a' and '1'
"ABBA" -> 2 # 'A' and 'B' each occur twice

Persistent Bugger.

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

For example (Input --> Output):

39 --> 3 (because 3*9 = 27, 2*7 = 14, 1*4 = 4 and 4 has only one digit)

999 --> 4 (because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2)

4 --> 0 (because 4 is already a one-digit number)