-1

This is my code atm:

puzzle = [
['1','1','1','1','1','1','1','1','1','1'],
['1','0','0','0','0','1','1','1','1','1'],
['1','1','0','1','0','0','0','0','0','0'],
['0','1','0','0','0','0','0','0','0','0'],
['0','0','1','0','1','0','1','0','0','0'],
['0','0','0','0','1','0','0','0','0','0'],
['1','0','0','0','1','0','1','0','1','0'],
['0','0','0','0','1','0','0','0','0','1'],
]

#30 1 s

counter1 = 0
counter2 = 0

for a in puzzle:
   b = ''.join(a)
   for search in b:
       if search == '1':
           counter1 = counter1 + 1
       if counter1 > counter2:
           counter2 = counter1

#output should be 3
print(counter1, counter2)

Now I am trying to loop vertically through my list. So I can find the highest amount of 1's after each other without a 0 interfering. Starting with counting from the top left. Only I haven't the slightest idea on how to do that. #output should be 3 Could someone help me out?

CaptainGuan
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2 Answers2

0

This is how I would go about it

puzzle_new = list(map(list, zip(*puzzle)))

puzzle_new will now look like the following:

[['1', '1', '1', '0', '0', '0', '1', '0'],
['1', '0', '1', '1', '0', '0', '0', '0'],
['1', '0', '0', '0', '1', '0', '0', '0'],
['1', '0', '1', '0', '0', '0', '0', '0'],
['1', '0', '0', '0', '1', '1', '1', '1'],
['1', '1', '0', '0', '0', '0', '0', '0'],
['1', '1', '0', '0', '1', '0', '1', '0'],
['1', '1', '0', '0', '0', '0', '0', '0'],
['1', '1', '0', '0', '0', '0', '1', '0'],
['1', '1', '0', '0', '0', '0', '0', '1']]

This will transpose the puzzle list. You can use the same code that you have for the rest of the logic on puzzle_new

Dharman
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Sahana Raguram
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0

Vectorized NumPy solution (if I understood your question correctly):

import numpy as np

puzzle = p = [
['1','1','1','1','1','1','1','1','1','1'],
['1','0','0','0','0','1','1','1','1','1'],
['1','1','0','1','0','0','0','0','0','0'],
['0','1','0','0','0','0','0','0','0','0'],
['0','0','1','0','1','0','1','0','0','0'],
['0','0','0','0','1','0','0','0','0','0'],
['1','0','0','0','1','0','1','0','1','0'],
['0','0','0','0','1','0','0','0','0','1'],
]

p = np.array(puzzle).astype(int)
np.max(np.cumsum(np.diff(np.cumsum(p, axis=0), axis=0), axis=0), axis=0)

which gives the maximum number of consecutive 1s in each of the columns:

array([3, 2, 1, 1, 4, 1, 3, 1, 2, 2])

Edit - solution for the vertically flattened array:

import numpy as np

puzzle = p = [
['1','1','1','1','1','1','1','1','1','1'],
['1','0','0','0','0','1','1','1','1','1'],
['1','1','0','1','0','0','0','0','0','0'],
['0','1','0','0','0','0','0','0','0','0'],
['0','0','1','0','1','0','1','0','0','0'],
['0','0','0','0','1','0','0','0','0','0'],
['1','0','0','0','1','0','1','0','1','0'],
['0','0','0','0','1','0','0','0','0','1'],
]
p = np.array(puzzle).astype(int).flatten(order='F')
[sum(i) for i in np.split(p, np.where(np.diff(p) == 1)[0]+1)]

which gives:

[3, 1, 1, 2, 1, 1, 1, 1, 1, 6, 2, 1, 1, 2, 2, 1, 2, 1]
alex
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  • It is a good solution only thing is that you have to start from the top with the counting. So the 2nd is for example 1 and the 5th is also 1 – CaptainGuan Nov 02 '21 at 11:44
  • @CaptainGuan it's good to add an expected output to the question – alex Nov 02 '21 at 11:49
  • @CaptainGuan take a look at the updated solution. Is that what you were looking for? – alex Nov 02 '21 at 12:14
  • Could you explain how you got 6 in the output? – CaptainGuan Nov 02 '21 at 12:20
  • @CaptainGuan you flatten the matrix from the top left side downwards (like you said), split it into groups of contunuous `1`s and `0`s and count the number of `1`s in each group. There is 6 `1`s in one of them. – alex Nov 02 '21 at 13:41