Given a string S.
You have to return another string such that even-indexed and odd-indexed characters of S are grouped and groups are space-separated (see sample below)

Note: 
0 is considered to be an even index. 
All input strings are valid with no spaces


input:
'CodeWars'

output
'CdWr oeas'

S[0] = 'C'
S[1] = 'o'
S[2] = 'd'
S[3] = 'e'
S[4] = 'W'
S[5] = 'a'
S[6] = 'r'
S[7] = 's'


Even indices 0, 2, 4, 6, so we have 'CdWr' as the first group
odd ones are 1, 3, 5, 7, so the second group is 'oeas'
And the final string to return is 'Cdwr oeas'

Enjoy.

Difficulty: introductory

Starter code:

def sort_my_string(s):
	

Solutions:
def sort_my_string(s):
    return '{} {}'.format(s[::2], s[1::2])

def sort_my_string(s):
    return s[::2] + ' ' + s[1::2]

def sort_my_string(S):
    return (S[0::2] + " " + S[1::2])

def sort_my_string(S):
    return str(S[0::2] + ' ' + S[1::2])

def sort_my_string(s):
    return f'{s[::2]} {s[1::2]}'

def sort_my_string(s):
    odd, even = [], []
    for i, char in enumerate(s):
        even.append(char) if i % 2 == 0 else odd.append(char)
    return "".join(even) + " " + "".join(odd)

def sort_my_string(s):
    l = list(s)
    return "".join(l[::2]+[" "]+l[1::2])

def sort_my_string(str):
    
    evenChar = str[::2]
    oddChar  = str[1::2]

    return evenChar + ' ' + oddChar

#end function sort_my_string


def sort_my_string(s):
    return '%s %s' % (s[::2], s[1::2])

sort_my_string=lambda w:w[::2]+' '+w[1::2]