def isValid(s):
    cts=Counter(s) #Count occurences of string
    ct2=Counter(cts.values()) #Count unique occurences of string occurences
    #if only one item in ct2 then all occurences are equal, 
    #if two items then check if diff is 1 and second most has a count of 1
    #Special case: if the second occurences of occurences equals 1, 
    #it means the second 1 occurrence can be removed to make a valid string.
    kc=ct2.most_common()
    return 'YES' if len(kc)==1 or (len(kc)==2 and \
     (((kc[1][0]-kc[0][0])==1 or kc[1][0]==1) and kc[1][1]==1)) else 'NO'