# O(mn) solution for python.

• ``````def findBlackPixel(self, picture, N):
if not picture: return 0
m, n = len(picture), len(picture[0])
# storing 'B' coordinates so the second pass is faster. Could be removed if memory is constrained.
black_coordinates = []
# storing # of 'B' in a row and a column. For all 'B' in the same column, we also need their row id,
# and defaultdict is a convenient way to store it.
row, col = [0] * m, collections.defaultdict(list)
for i in xrange(m):
for j in xrange(n):
if picture[i][j] == 'B':
black_coordinates.append( (i, j) )
row[i] += 1
col[j].append(i)
count = 0
for (i, j) in black_coordinates:
#  -------    rule 1    --------------- rule 2, makesure all rows are the same using set-------
if row[i] == N and len(col[j]) == N and len(set([tuple(picture[r]) for r in col[j]])) == 1:
count += 1
return count
``````

• It's not O(mn) but O(m2n2). Think of a completely black picture of size N×N. For each of the N2 coordinates, you'll do N2 work.

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