Python - Single Pass - O(MN) time - O(M+N) space - Pruned Search Space

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    Some ideas we use to form this solution:

    • Assume we see the first 'B' in a particular row at column j1.
      If we see another 'B' in the same row at column j2, then we can be certain that no cell on this row/column j1/column j2 can ever have a lonely pixel.
    • If there was exactly one 'B' in the row, we go through column j1 from the current row and below. Any 'B's we encounter mean that there can never be a lonely pixel on the rows that we see these pixels.

    Essentially we have a set of possible rows and columns that could potentially have lonely pixels. Every time we see 'B's, we see if it's a lonely pixel (or eliminate potential rows/columns in the process).

        def findLonelyPixel(self, picture):
            m, n = len(picture), len(picture[0])
            lonely_pixel_count = 0
            invalid_rows, invalid_cols = [False]*m, [False]*n
            for row in xrange(m):
                # If this row has already been ininvalidated, we move on to the next row.
                if invalid_rows[row]:
                count, col = 0, None
                for j in xrange(n):
                    if picture[row][j] == 'B':
                        count, col = count+1, j
                        # If we have more than one 'B' on this row, then every column that has a 'B' is ininvalid
                        if count>1:
                            invalid_cols[col] = True
                            invalid_cols[j] = True
                if count==1 and not invalid_cols[col]:
                    lonely = True
                    for i in xrange(row+1, m):
                        # If we see another 'B' on the rest of the column, every row with a 'B' is invalid.
                        if picture[i][col] == 'B':
                            invalid_rows[i] = True
                            lonely = False
                    if lonely:
                    invalid_cols[col] = True
            return lonely_pixel_count

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