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headache solution in Creative category for Find Sequence by simleo
# migrated from python 2.7
import re
checkio = lambda a: any([re.search(r'(\d)\1{3,}', ''.join(map(str, s))) for s in [row[:] for row in a] + [[r[i] for r in a] for i in range(len(a))] + [[a[i+k][i] for i in range(len(a)-k)] for k in range(len(a)-3)] + [[a[i][i+k] for i in range(len(a)-k)] for k in range(1, len(a)-3)] + [[a[len(a)-i-k-1][i] for i in range(len(a)-k)] for k in range(len(a)-3)] + [[a[len(a)-i-1][i+k] for i in range(len(a)-k)] for k in range(1, len(a)-3)]])
March 7, 2014
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