I'd call this a cleanup of jpconcerman's solution using list comprehension and a ternary operator for the result. A plain old string will suffice for the vowel list as well but I'm not sure if a list or set ends up being computationally cheaper. For 5 characters though I'd assume the difference is negligible and all three will pass all test cases without timing out.

def minion_game(string):
    # your code goes here
    vowels = 'AEIOU'
    
    kevin = sum([len(string) - i for i in range(len(string)) if string[i] in vowels])
    stuart = sum([len(string) - i for i in range(len(string)) if string[i] not in vowels])
    
    print('Draw' if kevin == stuart
        else f'Kevin {kevin}' if kevin > stuart
        else f'Stuart {stuart}')

Can I understand more why it is sufficient to check the occurences of each character to keep the score, and we do not need the generate the words to check for occurrences?

def minion_game(string): vowels = ('A', 'E', 'I', 'O', 'U') kevin_score = 0 stuart_score = 0 len_str = len(string)

for i in range(len_str):
    if string[i] in vowels:
        kevin_score += (len_str - i)
    else:
        stuart_score += (len_str - i)

if kevin_score > stuart_score:
    print(f'Kevin {kevin_score}')
elif stuart_score > kevin_score:
    print(f'Stuart {stuart_score}')
else:
    print('Draw')

if name == 'main': s = input() minion_game(s)

def minion_game(string): n = len(string) S = string.upper() # be robust to case vowels = set("AEIOU")

kevin = 0   # vowel player
stuart = 0  # consonant player

for i, ch in enumerate(S):
    if ch in vowels:
        kevin += n - i
    else:
        stuart += n - i

if kevin > stuart:
    print(f"Kevin {kevin}")
elif stuart > kevin:
    print(f"Stuart {stuart}")
else:
    print("Draw")