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    Time Complexity :

    def repeatedString(s:str, n:int):
        # Number of letter A in s
        n_of_a = s.count("a")
    
        #Lenght of s
        n_of_letters = len(s)
    
        # number of s based on n , rest of s if exist 
        number_of_substring , rest_of_substring = divmod(n,n_of_letters)
        
        # number of letter A in the rest of s
        rest = s[0:rest_of_substring].count("a")
        
        # so every substring there exist number of A Letter + rest if exist  
        result = (number_of_substring * n_of_a)  +  rest
    
        return result