This is the best ans :

for i in range(n): if string[i] in vowels: kevin += n - i else: stuart += n - i

if stuart > kevin: print("Stuart", stuart) elif kevin > stuart: print("Kevin", kevin) else: print("Draw")

This is the best ans :

for i in range(n): if string[i] in vowels: kevin += n - i else: stuart += n - i

if stuart > kevin: print("Stuart", stuart) elif kevin > stuart: print("Kevin", kevin) else: print("Draw")

The Minion Game – Efficient Approach

To solve The Minion Game, generating all substrings is not required and will cause performance issues for large strings.

The key idea is that each character contributes to multiple substrings based on its position:

If the character is a vowel, the score goes to Kevin

If it’s a consonant, the score goes to Stuart

The number of substrings starting at index i is len(string) - i

Using this logic, the problem can be solved in O(N) time with a single loop.

def minion_game(string): vowels = "AEIOU" kevin = 0 stuart = 0 n = len(string)

for i in range(n):
    if string[i] in vowels:
        kevin += n - i
    else:
        stuart += n - i

if stuart > kevin:
    print("Stuart", stuart)
elif kevin > stuart:
    print("Kevin", kevin)
else:
    print("Draw")

This approach avoids substring creation and works efficiently for all test cases.

def minion_game(string): # your code goes here stuart_score = 0 kevin_score = 0 score = 0 vowels = 'aeiou' for i in range(len(string)): score = len(string) - i if string[i].lower() in vowels: kevin_score += score else: stuart_score += score

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

def minion_game(string): # your code goes here a = len(string) k = 0 s = 0 for i in range(a): if string[i].lower() in ['a','e','i','o','u']: k = a-i + k else: s = s+ a-i if s > k: print((f'Stuart {s}')) elif k>s: print((f'Kevin {k}')) else: print("Draw")

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