Hardest Cryptarithmetic Puzzle

I have found two ways of doing this, one a shorter code length but longer solve time, and one with a longer code length, but shorter solver time

from re import sub
def solve(q):
    try:
        n = next((i for i in q if i.isalpha()))
    except StopIteration:
        return q if eval(sub(r'(^|[^0-9])0+([1-9]+)', r'\1\2', q)) else False
    else: 
        for i in (str(i) for i in range(10) if str(i) not in q):
            r = solve(q.replace(n,str(i)))
        if r:
            print(r)
        return

if __name__ == "__main__":
    query = "AA + BB == CC" #If you enter the problem here, it will solve it
    solve(query)

or

import itertools, time
sentences = [
    "AA + BB == CC" #If you enter the problem here, it will solve it
    ]
def show_solution(s):
    print(s)
    print('{' + ', '.join(['"'+ chr(i) + '"=>' + chr(trantab[i]) for i in trantab]) + '}\n')
for sentence in sentences:
    t1 = time.time()
    words = [ss for ss in sentence.split() if ss.isalpha()]
    letters = ''
    for w in words:
        for l in w:
            if l not in letters:
                letters += l        ## ELL not ONE
    letters = ''.join(sorted(letters))
    length = len(letters)
    for p in itertools.permutations(("0","1","2","3","4","5","6","7","8","9"), length):
        p2 = "".join(p)
        trantab = str.maketrans(letters, p2)
        ww = []
        lead_0 = False
        for w in words:
            w2 = w.translate(trantab)
            if w2[0] == "0":
                lead_0 = True
                break
            ww.append(int(w2))
        if not lead_0 and sum(ww[:-1]) == ww[-1]:
            show_solution('"{0}": {1:.2f} sec.'.format(sentence, time.time() - t1))
            break