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existence-of-a-substring-in-a-string-and-its-reverse.py

for i in xrange(len(s)-1): lookup[ord(s[i])-ord('a')][ord(s[i+1])-ord('a')] = True return any(lookup[ord(s[i+1])-ord('a')][ord(s[i])-ord('a')] for i in xrange(len(s)-1)) ...