# Short Python

• Mainly I group and count equal rows. Look for rows that appear N times and that have N black pixels. If you find one, add N for each of its black columns that doesn't have extra black pixels (in other rows).

``````def findBlackPixel(self, picture, N):
ctr = collections.Counter(map(tuple, picture))
cols = [col.count('B') for col in zip(*picture)]
return sum(N * zip(row, cols).count(('B', N))
for row, count in ctr.items()
if count == N == row.count('B'))``````

• I came up with some ideas based on your amazing solution in Pixel I. Just attached my solution in discussion. Thanks for your help!

• Input:
["WBWBBW",
"BWBWWB",
"WBWBBW",
"BWBWWB",
"WWWBBW",
"BWBWWB"]
3
why the except is 9 not 13?

• @Deathlok
Maybe you forgot the second requirement which is `"2. For all rows that have a black pixel at column C, they should be exactly the same as row R"`. For the input you mentioned, only 3 `"BWBWWB"` are "valid" to be consider for answer, so yes, it is 9 instead of 13.

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