String contains all the elements of a list

>>> all(x in 'tomato' for x in ['t','o','m','a'])
True
>>> all(x in 'potato' for x in ['t','o','m','a'])
False

def myfun(str,list):
   for a in list:
      if not a in str:
         return False
   return True

return true must be outside the for loop, not just after the if statement, otherwise it will return true just after the first letter has been checked. this solves your code's problem :)

For each letter you go through the list. So if the list is of length n and you have m letters, then complexity is O(n*m). And you may achieve O(m) if you preprocess the word.

def myfun(word,L):
    word_letters = set(word) #This makes the lookup `O(1)` instead of `O(n)`
    return all(letter in word_letters for letter in L)

Also, it's not a good practice to name variables as str and list as if you will need later to create list or use str, they will be shaded by your variables.

Some relevant information:

• all function

• set complexity

If you're not worried about repeat characters, then:

def myfunc(string, seq):
    return set(seq).issubset(string)

And, untested, if you do care about repeated characters, then maybe (untested):

from collections import Counter
def myfunc(string, seq):
    c1 = Counter(string)
    c2 = Counter(seq)
    return not (c2 - c1)