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First solution in Clear category for Find Sequence by Mateusz_Tomkowiak
def checkSequence(arr):
counter = 1
for i in range(len(arr)-1):
if arr[i] == arr[i+1]:
counter += 1
else:
counter=1
if counter>3:
return True
def checkio(matrix):
for row in matrix:
if(checkSequence(row)):
return True
for i in range(len(matrix)):
col = []
for j in range(len(matrix)):
col.append(matrix[j][i])
if(checkSequence(col)):
return True
for i in range(len(matrix) - 3):
diagonal1 = [matrix[j + i][j] for j in range(len(matrix) - i)]
if checkSequence(diagonal1): return True
diagonal2 = [matrix[j][j + i] for j in range(len(matrix) - i)]
if diagonal1 != diagonal2 and checkSequence(diagonal2): return True
diagonal3 = [matrix[len(matrix) - 1 - j - i][j] for j in range(len(matrix) - i)]
if checkSequence(diagonal3): return True
diagonal4 = [matrix[len(matrix) - 1 - j][j + i] for j in range(len(matrix) - i)]
if diagonal3 != diagonal4 and checkSequence(diagonal4): 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. 23, 2016
