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'

def isValid(s):
    # Write your code here
    holder = {}
    for char in s:
        if char in holder:
            holder[char] += 1
        else:
            holder[char] = 1
    val = list(holder.values())
    val.sort()
    test1 = set(val[1:])
    test2 = set(val[:-1])
    test2.add(val[-1]-1)
    result = "YES" if len(test1) == 1 or len(test2) == 1 else "NO"
    return result

def isValid(s):
    arr = [0] * 26
    for i in range(len(s)):
        arr[ord(s[i])-97]+=1
    arr.sort(reverse=True)
    if min(arr) == 0:
        arr= arr[:arr.index(0)]
    if len(arr) == 1:
        return "YES"
    mainValue = arr[1]
    firstValue = arr[0]
    lastValue = arr[-1]
    left = CheckIsAllSame(arr[0:len(arr)-1])
    if left and (lastValue == mainValue or lastValue == 1):
        return "YES"
    right = CheckIsAllSame(arr[1:len(arr)])
    if right and (firstValue == mainValue or firstValue == mainValue + 1):
        return "YES"
    return "NO"

def CheckIsAllSame(arr):
    for i in range(len(arr)-1):
        if arr[i] != arr[i+1]:
            return False
    return True