Lonely pixel I

Time: O(M N); Space: O(M + N); med*

Given a picture consisting of black and white pixels, find the number of black lonely pixels. The picture is represented by a 2D char array consisting of ‘B’ and ‘W’, which means black and white pixels respectively. A black lonely pixel is character ‘B’ that located at a specific position where the same row and same column don’t have any other black pixels.

Example 1:

Input: picture =

[
    ['W', 'W', 'B'],
    ['W', 'B', 'W'],
    ['B', 'W', 'W']
]

Output: 3

Explanation:

  • All the three ’B’s are black lonely pixels.

Note:

  • The range of width and height of the input 2D array is [1,500].

[1]:

class Solution1(object):
    def findLonelyPixel(self, picture):
        """
        :type picture: List[List[str]]
        :rtype: int
        """
        rows, cols = [0] * len(picture),  [0] * len(picture[0])
        for i in range(len(picture)):
            for j in range(len(picture[0])):
                if picture[i][j] == 'B':
                    rows[i] += 1
                    cols[j] += 1

        result = 0
        for i in range(len(picture)):
            if rows[i] == 1:
                for j in range(len(picture[0])):
                     result += picture[i][j] == 'B' and cols[j] == 1
        return result

[2]:

s = Solution1()
picture = [
    ['W', 'W', 'B'],
    ['W', 'B', 'W'],
    ['B', 'W', 'W']
]
assert s.findLonelyPixel(picture) == 3

[3]:

class Solution2(object):
    def findLonelyPixel(self, picture):
        """
        :type picture: List[List[str]]
        :rtype: int
        """
        return sum(col.count('B') == 1 == picture[col.index('B')].count('B') \
               for col in zip(*picture))

[4]:

s = Solution2()
picture = [
    ['W', 'W', 'B'],
    ['W', 'B', 'W'],
    ['B', 'W', 'W']
]
assert s.findLonelyPixel(picture) == 3