检查字符串是否具有相同频率的字符，删除或不删除 1 个字符

Question

我正在尝试编写代码来检查给定字符串是否满足以下条件：

1. 出现频率相同的所有字符
2. 如果需要，可以删除一个字符以实现上述条件

以下代码适用于某些测试用例，但不适用于其他用例，您能帮帮我吗？

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

试试这个 -

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