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Using Counter from collections and sorted. Less iterations solution in Clear category for The Most Wanted Letter by jcfernandez
def checkio(string):
from collections import Counter
from string import ascii_letters
string = sorted(
sorted(
# Count how many times does a letter repeats
Counter(
# Avoid counting anything but letters
char for char in string.lower() if char in ascii_letters
).most_common(),
# Counter(...).most_common() returns a list of tuples containing each letter and its counting
# We 1st sort the list of tuples alphabetically
key=lambda char_count_tpl: char_count_tpl[0]
),
# Then we re-sort by letter counting, from higher to lower
key=lambda char_count_tpl: char_count_tpl[1], reverse=True
)
# Get the letter with the most counting
# We return only the letter
return string[0][0] if string else ''
if __name__ == '__main__':
print("Example:")
print(checkio("Hello World!"))
# These "asserts" using only for self-checking and not necessary for auto-testing
assert checkio("Hello World!") == "l", "Hello test"
assert checkio("How do you do?") == "o", "O is most wanted"
assert checkio("One") == "e", "All letter only once."
assert checkio("Oops!") == "o", "Don't forget about lower case."
assert checkio("AAaooo!!!!") == "a", "Only letters."
assert checkio("abe") == "a", "The First."
print("Start the long test")
assert checkio("a" * 9000 + "b" * 1000) == "a", "Long."
print("The local tests are done.")
June 22, 2019
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