You need to write a function, that returns the first non-repeated character in the given string.

For example for string `"test"` function should return `'e'`.  
For string `"teeter"` function should return `'r'`.  
  
If a string contains all unique characters, then return just the first character of the string.  
Example: for input `"trend"` function should return `'t'`  
  
You can assume, that the input string has always non-zero length.

If there is no repeating character, return `null` in JS or Java, and `None` in Python.

Difficulty: introductory

Starter code:

def first_non_repeated(s):
	

Solutions:
def first_non_repeated(s):
    return next((c for c in s if s.count(c) == 1), None)

def first_non_repeated(s):
    for c in s:
        if s.count(c) == 1: return c

from collections import Counter

def first_non_repeated(s):
    return next((k for k,v in Counter(s).items() if v==1), None)

def first_non_repeated(string):
    for i in range(len(string.upper())):
        e = string[i]
        if (string.lower().count(e) == 1):
            return e
    return None

def first_non_repeated(s):
    print(s)
    return next((i for i in s if s.count(i) == 1), None)

def first_non_repeated(s):
    for a in s:
        if s.count(a) == 1:
            return a
    return None

def first_non_repeated(s):
    return next((x for x in s if s.count(x) == 1), None)

def first_non_repeated(s):
    for character in s:
        if s.count(character) == 1:
            return character
            break

def first_non_repeated(s):
    for el in s:
        if s.count(el) == 1:
            return el

def first_non_repeated(s):
    sets = unique, repeated = set(), set()
    for c in s: sets[c in unique].add(c)
    return next((c for c in s if c not in repeated), None)