The challenge

Consider the word "abode". We can see that the letter a is in position 1 and b is in position 2. In the alphabet, a and b are also in positions 1 and 2. Notice also that d and e in abode occupy the positions they would occupy in the alphabet, which are positions 4 and 5.

Given an array of words, return an array of the number of letters that occupy their positions in the alphabet for each word. For example,

solve(["abode","ABc","xyzD"]) = [4, 3, 1]

See test cases for more examples.

Input will consist of alphabet characters, both uppercase and lowercase. No spaces.

The solution in Python code

Option 1:

from operator import eq
from string import ascii_lowercase

def solve(strings):
    return [sum(map(eq, s.lower(), ascii_lowercase)) for s in strings]

Option 2:

def solve(arr):
    return [sum(i == ord(c) - ord('A') for i, c in enumerate(s.upper())) for s in arr]

Option 3:

def solve(words):
  return [sum(a==b for a, b in zip(w.lower(), 'abcdefghijklmnopqrstuvwxyz')) for w in words]

Test cases to validate our solution

test.it("Basic tests")
test.assert_equals(solve(["abode","ABc","xyzD"]),[4,3,1])
test.assert_equals(solve(["abide","ABc","xyz"]),[4,3,0])
test.assert_equals(solve(["IAMDEFANDJKL","thedefgh","xyzDEFghijabc"]),[6,5,7])
test.assert_equals(solve(["encode","abc","xyzD","ABmD"]),[1, 3, 1, 3])