Some require counting all valid subarrays. By maintaining a valid window and counting all valid subarrays ending at each position, we can reduce an O (n²) approach to an efficient O (n) solution.
Updating the top, bottom, left, and right pointers after each traversal provides a clean and efficient way to process the matrix layer by layer. #LeetCode #30DaysOfLeetCode #Python #DSA ...
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