Main idea: each time find ch == ‘c’ check top of stack of they are ‘a’,’b’ in order, then pop them off time O(N) space O(N)

class Solution:
    def isValid(self, s: str) -> bool:
        # checki if s is potential candidate   
        counter  = collections.Counter(s)
        if len(set(counter.values())) != 1:
            return False
        if s[0] != 'a':
            return False
        st = []
        i = 0
        n = len(s)
        while i < n:
            if s[i] == 'c' :
                if len(st) >= 2 and st[-1] == 'b' and st[-2] == 'a':
                    st.pop()
                    st.pop()
                    i += 1 
                    continue
            st.append(s[i])
            i += 1 
        return True if len(st) == 0 else False

faster solution

class Solution:
    def isValid(self, S):
        S2 = ""
        while S != S2:
            S, S2 = S.replace("abc", ""), S
        return S == ""