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def is_palindromic(text):
index = 0
while index < len(text) / 2:
if text[index] != text[len(text) - index - 1]:
return False
index += 1
return True
def yield_substring(text, ele):
# return substring of given text with ele as the first & last char
indexes = [i for i in range(len(text)) if text[i] == ele]
for i in range(len(indexes) - 1):
for j in range(i + 1, len(indexes)):
yield text[indexes[i] : indexes[j] + 1]
def longest_palindromic(text):
text_elements = set(text)
palindromic = text[0]
for i in text_elements:
for j in yield_substring(text, i):
if is_palindromic(j):
if len(j) == len(palindromic) and text.find(j) < text.find(palindromic):
palindromic = j
elif len(j) > len(palindromic):
return palindromic
if __name__ == '__main__': # pragma: no cover
assert longest_palindromic("artrartrt") == "rtrartr", "The Longest"
assert longest_palindromic("abacada") == "aba", "The First"
assert longest_palindromic("aaaa") == "aaaa", "The A"
KenMercusLai /
checkio
