My first instinct was to brute force it by checking every possible top-left and bottom-right pair, which works but is slow.
Clarify the problem constraints (matrix dimensions, color alphabet size) and discuss brute-force O(n^4) checking all pairs of rows and columns. Then optimize by grouping cells by color and for each color, count pairs of rows that share at least two columns of that color, using combinatorial counting. Finally, analyze time and space complexity and consider edge cases.
Pro tip: Mention that the problem can be solved in O(n^2 * m) time by iterating over column pairs and using a hash map to count row pairs with matching colors, which is more efficient than the naive O(n^4) approach. Also, discuss how to handle large matrices with memory constraints.
Ask about matrix dimensions, color representation, and whether squares can be of any size (including 1x1). Confirm that corners must be distinct cells.
Explain the naive O(n^4) solution: iterate over all pairs of rows and columns, check if the four corners have the same color, and count.
Propose grouping cells by color. For each color, collect all positions. Then for each pair of rows, count how many columns have that color in both rows; if count >= 2, add C(count,2) to the total.
Analyze time and space complexity of both approaches. The optimized approach can be O(n^2 * m) or O(n * m^2) depending on implementation, and space O(n*m) for storing positions.
Discuss edge cases: empty matrix, 1x1 matrix, all cells same color, all cells different colors. Suggest testing with small matrices and verifying against brute-force.
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