Fix: Ensure all skills are tracked as files, not submodules
This commit is contained in:
@@ -0,0 +1,30 @@
|
||||
# HumanEval/117
|
||||
# Loki Mode Multi-Agent Solution
|
||||
# Attempts: 1
|
||||
# Passed: True
|
||||
|
||||
def select_words(s, n):
|
||||
"""Given a string s and a natural number n, you have been tasked to implement
|
||||
a function that returns a list of all words from string s that contain exactly
|
||||
n consonants, in order these words appear in the string s.
|
||||
If the string s is empty then the function should return an empty list.
|
||||
Note: you may assume the input string contains only letters and spaces.
|
||||
Examples:
|
||||
select_words("Mary had a little lamb", 4) ==> ["little"]
|
||||
select_words("Mary had a little lamb", 3) ==> ["Mary", "lamb"]
|
||||
select_words("simple white space", 2) ==> []
|
||||
select_words("Hello world", 4) ==> ["world"]
|
||||
select_words("Uncle sam", 3) ==> ["Uncle"]
|
||||
"""
|
||||
if not s:
|
||||
return []
|
||||
|
||||
vowels = "aeiouAEIOU"
|
||||
result = []
|
||||
|
||||
for word in s.split():
|
||||
consonant_count = sum(1 for char in word if char.isalpha() and char not in vowels)
|
||||
if consonant_count == n:
|
||||
result.append(word)
|
||||
|
||||
return result
|
||||
Reference in New Issue
Block a user