Return an output string that translates an input string `s`/`$s` by replacing each character in `s`/`$s` with a number representing the number of times that character occurs in `s`/`$s` and separating each number with the character(s) `sep`/`$sep`.

Difficulty: introductory

Starter code:

def freq_seq(s, sep):
	

Solutions:
def freq_seq(s, sep):
    return sep.join([str(s.count(i)) for i in s])


from collections import Counter

def freq_seq(s, sep):
    freq = Counter(s)
    return sep.join(str(freq[c]) for c in s)


from collections import Counter

def freq_seq(string, separator):
    count = Counter(string)
    return separator.join(str(count[symbol]) for symbol in string)


def freq_seq(s, sep):
    return sep.join(str(s.count(c)) for c in s)


from collections import Counter 

def freq_seq(s, sep):
    dic = Counter(s) 
    result = ''
      
    for char in s:
        for key, value in list(dic.items()):
            if char == key:
                result  += str(value)+sep
    
    
    return result[:-1]


def freq_seq(s, sep):
    return sep.join([str(s.count(c)) for c in s])


from collections import Counter

def freq_seq(string, sep):
    counter = Counter(string)
    return sep.join(str(counter[char]) for char in string)

def freq_seq(s, sep):
    numero = len(s)
    conta = 0
    contatore = 0
    contatore1 = 0
    lista = []
    while conta != numero:
        z = str(s[conta])
        while contatore1 < numero:
            if z == s[contatore1]:
                contatore += 1
            contatore1 += 1
        contatore1 = 0
        lista.append(str(contatore))
        contatore = 0
        conta += 1
    x = sep.join(lista)
    return x

    


def freq_seq(s, sep):
    return sep.join([str(s.count(letter)) for letter in s])

def freq_seq(s, sep):
    freq = {}
    newStr = []
    for i in s:
        if freq.get(i) == None:
            freq[i] = 1
        else:
            freq[i] += 1
    for i in s:
        newStr.append(str(freq[i]))
    return sep.join(newStr)