My first instinct was brute force: check every possible top-left corner, every possible size.
Start by clarifying the problem: count all k×k submatrices where all cells are the same color, for any k. Then propose an efficient dynamic programming solution that computes the largest monochromatic square ending at each cell, and use that to count all smaller squares. Discuss time and space complexity, and consider trade-offs between different approaches.
Pro tip: Mention that the DP approach can be optimized to O(1) space per row if only the count is needed, but be prepared to explain the trade-off between space and code clarity. Also, clarify whether overlapping squares are counted separately (they usually are).
Confirm that squares of any size k≥1 are counted, including 1×1, and that overlapping squares are counted multiple times. Ask if the grid can be large and if there are constraints on time/space.
Let dp[i][j] be the side length of the largest monochromatic square with bottom-right corner at (i,j). Base case: dp[i][j] = 1 if cell is valid (always). Recurrence: if grid[i][j] == grid[i-1][j] == grid[i][j-1] == grid[i-1][j-1], then dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]); else dp[i][j] = 1.
For each cell, the number of monochromatic squares ending at that cell is dp[i][j]. Sum dp[i][j] over all cells to get the total count. Explain why this works: if the largest square ending at (i,j) has side L, then there are exactly L squares (of sizes 1 to L) ending at that cell.
Time complexity: O(m*n) where m and n are grid dimensions. Space complexity: O(m*n) for the DP table, but can be optimized to O(n) by keeping only the previous row. Discuss trade-offs.
Handle empty grid, single row/column, and all same color. Mention that if the grid is very large, we might use a rolling array to save space. Also, note that the DP can be computed in-place if we don't need the original grid.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.