检查字符串是否具有相同频率的字符，无论是否删除 1 个字符

Question

I'm trying to write a code to check if a given string satisfies the following conditions:

1. all the characters occurring in the same frequency
2. one character can be removed if necessary to achieve the above condition

the following code works for some test cases but not for others, can u please help me?

d=0
r=0
def isValid(s):
    # Write your code here
    c =Counter(s) #count no occurences of chars in string
    q=0
    v=c.values() 
    val=list(v)  # convert it to list 
    res=Counter(val)  #count no of equal occurences
    d=res.values() 
    dat=list(d) 
    
    if len(dat)==1:
        r='YES'
        return(r)
    elif len(dat)>1:
        q=dat[1]-1
             
    if q==0:
        r='YES'
        return(r)
    else:
        r='NO'
        return(r)   ```

Answer 1

Try this -

from collections import Counter

s = 'AABBCCDD' # Just a EXAMPLE

def check(x):
    z = []
    for y in s:
        z.append(y)

    a = Counter(z)
    b = dict(a)
    c = list(b.values())
    check_list = []
    for ele in c:
        if ele % 2 != 0:
            check_list.append(ele)
    
    if len(check_list) > 1:
        return False

def isValid(s):
    a = Counter(s)
    b = dict(a)
    c = list(b.values())
    y = []
    for x in c:
        y.append(x)
        if x != y[0]:
            if check(x) == False:
                print('Not Valid')
                return
            
        
    print('Valid')
 
isValid(s) # Calling the function