Solution Explanation: Counting and Enumeration

This problem can be efficiently solved using a counting approach combined with enumeration. The core idea is to first count the number of black pixels ('B') in each row and each column. Then, we iterate through the matrix again. If a pixel is black and its row and column counts are both 1 (meaning it's the only black pixel in that row and column), we increment the count of lonely pixels.

Algorithm:

1. Initialization:

   • Create two arrays, rows and cols, to store the counts of black pixels in each row and column, respectively. Initialize all elements to 0. The size of rows will be the number of rows in the picture, and the size of cols will be the number of columns.

2. Counting Black Pixels:

   • Iterate through the input picture matrix. For each pixel:
     • If the pixel is 'B', increment the corresponding element in rows (for the row) and cols (for the column).

3. Counting Lonely Pixels:

   • Iterate through the picture matrix again. For each pixel:
     • If the pixel is 'B' and rows[i] (the count of black pixels in its row) is 1 and cols[j] (the count of black pixels in its column) is 1, then increment the ans (the count of lonely pixels).

4. Return Result:

   • Return the final count ans.

Time Complexity: O(m*n) - We iterate through the matrix twice, where 'm' is the number of rows and 'n' is the number of columns.

Space Complexity: O(m+n) - We use two arrays (rows and cols) whose sizes are proportional to the number of rows and columns in the input matrix.

Code Examples (with explanations inline):

The provided code examples in Python, Java, C++, Go, and TypeScript all follow this algorithm. Let's highlight the key parts of the Python solution:

class Solution:
    def findLonelyPixel(self, picture: List[List[str]]) -> int:
        rows = [0] * len(picture)  # Initialize row counts
        cols = [0] * len(picture[0]) # Initialize column counts
 
        # Count black pixels in each row and column
        for i, row in enumerate(picture):
            for j, x in enumerate(row):
                if x == "B":
                    rows[i] += 1
                    cols[j] += 1
 
        ans = 0 # Initialize lonely pixel count
        # Count lonely pixels
        for i, row in enumerate(picture):
            for j, x in enumerate(row):
                if x == "B" and rows[i] == 1 and cols[j] == 1:
                    ans += 1
 
        return ans

The other language examples follow the same structure, just using their respective syntax and data structures. The key is the two counting loops and the final loop that checks for the lonely pixel condition.