Solution Explanation for Lonely Pixel II

This problem asks to find the number of "lonely" black pixels in a matrix. A black pixel is lonely if:

1. Both its row and column contain exactly target black pixels.
2. All rows containing a black pixel in its column are identical.

The solution employs a counting approach to efficiently identify these lonely pixels. Let's break down the steps and analyze the time and space complexity.

Approach:

1. Initialization:

   • rows: An array to store the count of black pixels in each row.
   • g: An adjacency list (using a dictionary/map in Python/Java/C++/Go or an array of arrays in TypeScript). g[j] stores the row indices of all black pixels in column j.

2. Counting Black Pixels:

   • The code iterates through the matrix. For each black pixel 'B' at (i, j), it increments rows[i] and adds i to g[j]. This efficiently counts black pixels per row and builds the adjacency list.

3. Checking for Lonely Pixels:

   • The code iterates through each column j.
   • Skip if needed: If g[j] is empty (no black pixels in this column) or the number of black pixels in the first row with a black pixel in this column (rows[g[j][0]]) is not equal to target, it skips to the next column. This is an optimization to avoid unnecessary checks.
   • Identical Rows Check: It checks if all rows in g[j] are identical to the first row g[j][0]. If the lengths of g[j] and rows[g[j][0]] are different, the rows are not identical, and the ok flag is set to 0. If they are the same, the ok flag is temporarily set to target. If any row in g[j] is different from the first one, ok is set to 0.
   • Add to Answer: If all rows are identical and contain target black pixels, it adds target to the ans (the count of lonely pixels).

4. Return: The function returns ans, the total number of lonely black pixels.

Time Complexity Analysis:

• The initial counting of black pixels takes O(m * n) time, where 'm' is the number of rows and 'n' is the number of columns.
• The nested loop checking for identical rows within each column has a worst-case complexity of O(m * n * m) = O(m² * n). This is because, in the worst case, all the rows might need to be compared, which happens when the number of black pixels in a column approaches m.

Therefore, the overall time complexity is dominated by the nested loop and is O(m² * n).

Space Complexity Analysis:

• rows array takes O(m) space.
• g (adjacency list) in the worst case could store all the row indices for each column, resulting in O(m * n) space.

Hence, the overall space complexity is O(m * n).

Code Examples (with slight improvements for clarity):

The provided code examples are mostly efficient. A minor improvement could be explicitly checking the row lengths for identical rows checks before the comparison. This will improve the efficiency of comparison in many cases.

In summary: The solution provides an efficient approach to solve the Lonely Pixel II problem, with a time complexity of O(m² * n) and a space complexity of O(m * n). The given code examples effectively implement this approach in various programming languages.