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Regex solution in Speedy category for Find Sequence by drakes00
import re
def checkLines(lines):
return any([re.match(r".*(\d)\1{3,}.*", "".join(map(str, l))) for l in lines])
def getDiags(matrix):
return [[matrix[i+start][i] for i in range(len(matrix)-start)]+[matrix[i][i+start] for i in range(len(matrix)-start)] for start in range(len(matrix)-3)]
def mirror(matrix):
return [line[::-1] for line in matrix]
def checkio(matrix):
if checkLines(matrix):
return True
if checkLines(list(zip(*matrix))):
return True
if checkLines(getDiags(matrix)):
return True
if checkLines(getDiags(mirror(matrix))):
return True
return False
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
assert checkio([
[1, 2, 1, 1],
[1, 1, 4, 1],
[1, 3, 1, 6],
[1, 7, 2, 5]
]) == True, "Vertical"
assert checkio([
[7, 1, 4, 1],
[1, 2, 5, 2],
[3, 4, 1, 3],
[1, 1, 8, 1]
]) == False, "Nothing here"
assert checkio([
[2, 1, 1, 6, 1],
[1, 3, 2, 1, 1],
[4, 1, 1, 3, 1],
[5, 5, 5, 5, 5],
[1, 1, 3, 1, 1]
]) == True, "Long Horizontal"
assert checkio([
[7, 1, 1, 8, 1, 1],
[1, 1, 7, 3, 1, 5],
[2, 3, 1, 2, 5, 1],
[1, 1, 1, 5, 1, 4],
[4, 6, 5, 1, 3, 1],
[1, 1, 9, 1, 2, 1]
]) == True, "Diagonal"
Oct. 16, 2014
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