Introduction

Problem: Given a list of words, return the list of numbers which represents the number of letters in that word that is the same as its position in the alphabet. Source

Final solution:

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

First attempt:

def solve(arr):
    letters = "abcdefghijklmnopqrstuvwxyz"
   
    def f(word):
        c = 0 
        for (i, j) in enumerate(word):
            if i > 25: return c
            if letters[i] == j: 
                c+=1
        return c
         
    arr = map(lambda x: x.lower(), arr)
    res = list(map(f, arr))
    return res

solve(["encode","abc","xyzD","ABmD"]) // 1, 3, 1, 4
solve(["abode","ABc","xyzD"]) // 4, 3, 1

The lessons

Get rid of lambdas and map 

arr = map(lambda x: x.lower(), arr)
# can always be replaced by
arr = [x.lower for x in arr]

A counter can be replaced with sum 

c = 0
...
if cond: c+=1
...
# Can be replaced by
sum(cond for ...)

# e.g.
def f(word):
    return sum(letters[i] == j for (i,j) in enumerate(word))

Trick for finding distance of letters 

# Instead of hard coding letters, or numbers
letters = "abcd..."
letters[i] == j
# Replace it with
i == Ord(j) - Ord("a")

Conclusion

We have seen how we can use list comprehension to simplify our problem.