The problem
Take into account the phrase "abode". We will see that the letter a is in place 1 and b is in place 2. Within the alphabet, a and b are additionally in positions 1 and 2. Discover additionally that d and e in abode occupy the positions they might occupy within the alphabet, that are positions 4 and 5.
Given an array of phrases, return an array of the variety of letters that occupy their positions within the alphabet for every phrase. For instance,
resolve(["abode","ABc","xyzD"]) = [4, 3, 1]
See check instances for extra examples.
Enter will include alphabet characters, each uppercase and lowercase. No areas.
The answer in Python code
Choice 1:
from operator import eq
from string import ascii_lowercase
def resolve(strings):
return [sum(map(eq, s.lower(), ascii_lowercase)) for s in strings]
Choice 2:
def resolve(arr):
return [sum(i == ord(c) - ord('A') for i, c in enumerate(s.upper())) for s in arr]
Choice 3:
def resolve(phrases):
return [sum(a==b for a, b in zip(w.lower(), 'abcdefghijklmnopqrstuvwxyz')) for w in words]
Take a look at instances to validate our resolution
check.it("Fundamental assessments")
check.assert_equals(resolve(["abode","ABc","xyzD"]),[4,3,1])
check.assert_equals(resolve(["abide","ABc","xyz"]),[4,3,0])
check.assert_equals(resolve(["IAMDEFANDJKL","thedefgh","xyzDEFghijabc"]),[6,5,7])
check.assert_equals(resolve(["encode","abc","xyzD","ABmD"]),[1, 3, 1, 3])
