At first I wanted to say it is best practice but it takes a lot to understand the tricks that make it work. It is extremelly clever though but I would not say this is easy to modify if the code has to evolve. The changing direction condition is particularly difficult to read and this is probably why you added the comments but it could be a lot clearer by splitting into multiple "elif" in my opinion.

I really like the self.moving and it is clever the way you use it in areSomeGuysWaitingSameWay_AtThatFloor and areSomeGuysWaitingOtherWay_AtThatFloor.

The function should also return false if the string has a length of less than 3 letters or fewer than 3 alphabetical characters.

Sample test error message for 'bob' is wrong, it's testing for true instead of for false.

The sample test for test "aa" and "bob" expect false when they are in fact plaindromes.

actually "T[space]T"

# You can generalise the function to only have one handling number by slice of 3 digit and call it recursively (here is a qwick exemple) 
convertionTable={0:'zero ',1:'one ',2:'two ',3:'three ',4:'four ',5:'five ',6:'six ',7:'seven ',8:'eight ',9:'nine ',
10:'ten ',11:'eleven ',12:'twelve ',13:'thirteen ',14:'fourteen ',15:'fifteen ',16:'sixteen ',17:'seventeen ',18:'eighteen ',19:'nineteen ',
20:'twenty ',30:'thirty ',40:'forty ',50:'fifty ',60:'sixty ',70:'seventy ',80:'eighty ',90:'ninety '}

def convert(num, depth=0):
    string = ""
    rest = num

    if(num > 999):
        string += convert(int(num/1000), depth+1)
        rest = rest%1000

    cent = int(rest/100)
    if(cent != 0):
        string += convertionTable[cent]+"hundred "
        rest = num%100

    if(rest and string and not depth):
      string += "and "

    diz = int(rest/10)
    if(diz > 1):
        string += convertionTable[diz*10]
        rest = rest%10
    elif(diz):
        string += convertionTable[rest]
        rest = 0

    unit = rest
    if(not string):
        string += convertionTable[unit]
    elif(unit > 0):
        string += convertionTable[unit]
    rest = 0

    if(depth == 2):
        string += "million "
    elif(depth == 1):
        string += "thousand "
    else:
        string = string.strip()

    return string


print(convert(175684354))
print(convert(515096))
print(convert(1015))
print(convert(1000))  
print(convert(999))
print(convert(0))
print(convert(9))
print(convert(10))
print(convert(11))
print(convert(20))
print(convert(21))
print(convert(32))
print(convert(43))
print(convert(76))

Duration of execution is a criteria but this shouldn't be THE criteria to determine the best solution. The best solution should be a fine balance between complexity(related to execution time), readability and best practice.

Great,thank you！

Approved

Needed the new comment to resolve the issue.

Needed the new comment to resolve the issue.

Thanks for your findings. I have reduce the number for the last case then. Hope the issue never happen after this update.

I think that in your case the comments don't make the code easier to read and understand otherwise your solution is clean. Maybe just less efficient since you used a dict instead of a set.
Also you did the same "mistake" i did, you can get rid of the else statment because you always want to add the checked value.

had two issue before completing the kata with exactly the same code:

first attempt I got:
Signal:
Process exited prematurely with a SIGKILL signal.

second one i got this:
STDERR:
Traceback:
in
File "./frameworks/python/cw-2.py", line 22, in assert_equals
equals_msg = "{0} should equal {1}".format(repr(actual), repr(expected))
MemoryError

finally on the third attempt it passes