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First solution in Clear category for Common Words by Rafal.U
def checkio(first, second):
first = first.split(",")
second = second.split(",") # stworzenie tablicy stringów z wyrazów oddzielonych przecinkami
tab_words = []
wynik = ""
for i in first:
for j in second:
if i == j: # porównanie tablic stringów
tab_words.append(i) # dodanie do tablicy wpólnych slów z porównywanych tablic stringów
x = sorted(tab_words) # sortowanie tablicy stringów
s = ","
wynik = s.join(x) # połączenie posortowanych stringów w jedna sekwencje z przecinkami odzielajacymi słowa
return wynik
return ""
#These "asserts" using only for self-checking and not necessary for auto-testing
if __name__ == '__main__':
assert checkio("hello,world", "hello,earth") == "hello", "Hello"
assert checkio("one,two,three", "four,five,six") == "", "Too different"
assert checkio("one,two,three", "four,five,one,two,six,three") == "one,three,two", "1 2 3"
Nov. 13, 2016