Написать функцию, которая получает строку и преобразует ее

string = "hello world"
print(''.join(['(' if string.count(string[i:i+1]) > 1 else ')' for i in range(len(string)-1)]))

import re
 
s = "hello world"
print(''.join(')' if i in re.findall(r'(.)(?=.*\1)', s) else '(' for i in s))

>>> s = "hello world"
>>> def duplicate_encode(s:str) -> str:
    m={w: s.count(w) for w in s}
    for w, cnt in m.items():
        s = s.replace(w, "(" if cnt ==1 else ")")
    return s
 
>>> duplicate_encode("hello world")=="(()))(()()("
True
>>> def duplicate_encode(s:str) -> str:
    m = dict()
    for w in s:
        m.setdefault(w,0)
        m[w]+=1
    for w, cnt in m.items():
        s = s.replace(w, "(" if cnt ==1 else ")")
    return s
 
>>> duplicate_encode("hello world")=="(()))(()()("
True

from collections import Counter
 
 
def duplicate_encode(s: str) -> str:
    c = Counter(s.lower())
    return "".join("(" if c[x] == 1 else ")" for x in s)
   
assert duplicate_encode("hello world") == "(()))(()()("

string = "hello world"
print(''.join([')' if string.count(string[i:i+1]) > 1 else '(' for i in range(len(string))]))

print(''.join([')' if string.count(i) > 1 else '(' for i in string]))

print(''.join(')' if string.count(i) > 1 else '(' for i in string))

s = "hello world"  * 1000000

from collections import Counter
import re
 
 
def duplicate_encode(s:str) -> str:
    '''
    %timeit -r 2 -n 2 duplicate_encode(s)
    717 ms ± 3.88 ms per loop (mean ± std. dev. of 2 runs, 2 loops each)
    '''
    m = set(s)
    for w in m:
        s = s.replace(w, "(" if s.count(w) == 1 else ")")
    return s
 
 
 
def duplicate_encode1(s:str) -> str:
    '''%timeit -r 2 -n 2 duplicate_encode1(s)
    1.36 s ± 17.7 ms per loop (mean ± std. dev. of 2 runs, 2 loops each)
    '''
    m = Counter(s)
    for w in m:
        s = s.replace(w, "(" if m[w] == 1 else ")")
    return s
 
 
def duplicate_encode2(s:str) -> str:
    '''
    %timeit -r 2 -n 2 duplicate_encode2(s)
    3.67 s ± 16.3 ms per loop (mean ± std. dev. of 2 runs, 2 loops each
    '''
    m = dict()
    for w in s:
        m.setdefault(w,0)
        m[w]+=1
    for w, cnt in m.items():
        s = s.replace(w, "(" if cnt ==1 else ")")
    return s
 
 
 
def duplicate_encode3(s:str) -> str:
    '''%timeit -r 2 -n 2 duplicate_encode3(s)
    5 s ± 12.2 ms per loop (mean ± std. dev. of 2 runs, 2 loops each)
    '''
    m = Counter(s)
    s = "".join("(" if m[w] ==1 else ")" for w in s )
    return s
 
 
def duplicate_encode4(s:str) -> str:
    '''
    %timeit -r 2 -n 2 duplicate_encode4(s)
    6.77 s ± 61.7 ms per loop (mean ± std. dev. of 2 runs, 2 loops each)
    '''
    m = dict()
    for w in s:
        m.setdefault(w,0)
        m[w]+=1
    s = "".join("(" if m[w] ==1 else ")" for w in s )
    return s
 
 
def duplicate_encode5(s:str) -> str:
    '''
    %timeit -r 2 -n 2 duplicate_encode5(s)
    даже не пробуйте :-)
    '''
    return ''.join(')' if s.count(w) > 1 else '(' for w in s)
 
 
def duplicate_encode6(s:str) -> str:
    '''
    %timeit -r 2 -n 2 duplicate_encode6(s)
    даже не пробуйте :-)
    '''
    return ''.join(')' if i in re.findall(r'(.)(?=.*\1)', s) else '(' for i in s)