Find the number of possible combinations for groups of 3 substrings with the same amount of letter `a`

substring_groups.py

from collections import Counter

def solution_aux(S, groupSize, groups, numberOfAs):
    print(groups, S, numberOfAs)

    if len(groups) == 3:
        if len(S) == 0:
            print('count')
            return 1

    if len(groups) > 3:
        return 0

    if numberOfAs > groupSize:
        return 0

    if len(S) == 0:
        return 0

    result = 0
    if S[0] == 'a':
        numberOfAs += 1
        if numberOfAs < groupSize:
            groups[-1] += S[0]
            result += solution_aux(S[1:], groupSize, groups[:], numberOfAs)
        else:
            groups[-1] += S[0]
            result += solution_aux(S[1:], groupSize, groups[:], numberOfAs)

            if numberOfAs > groupSize:
                groups[-1] = groups[-1][:-1]
                groups.append(S[0])
                result += solution_aux(S[1:], groupSize, groups[:], 1)
    else:
        if numberOfAs < groupSize:
            groups[-1] += S[0]
            result += solution_aux(S[1:], groupSize, groups[:], numberOfAs)
            groups[-1] = groups[-1][:-1]
            groups.append(S[0])
            result += solution_aux(S[1:], groupSize, groups[:], numberOfAs)
        else:
            groups[-1] += S[0]
            result += solution_aux(S[1:], groupSize, groups[:], numberOfAs)
            groups[-1] = groups[-1][:-1]
            groups.append(S[0])
            result += solution_aux(S[1:], groupSize, groups[:], 0)

    return result



def solution(S):
    letters = Counter(S)

    if letters['a'] % 3 != 0:
        return 0

    groupSize = letters['a'] // 3

    return solution_aux(S[1:], groupSize, [S[0]], int(S[0] == 'a'))

x  = solution('bbbbb')
print(x)