The classic all-ones squares DP problem but generalized to multiple colors.
Use dynamic programming to compute the size of the largest monochromatic square ending at each cell, then sum these sizes to count all valid squares. Alternatively, for each cell, expand outward to find all squares, but DP is more efficient. Explain the DP recurrence and how it avoids redundant checks.
Pro tip: Clarify the definition of 'colored cells' and whether colors are given as integers or strings; also discuss the trade-off between time and space complexity, and mention that the DP solution can be optimized to O(n) space if only the count is needed.
Confirm the grid dimensions, color representation, and that squares must be contiguous and axis-aligned. Ask if overlapping squares are counted separately.
Let dp[i][j] be the side length of the largest monochromatic square with bottom-right corner at (i, j). The number of squares ending at (i, j) is exactly dp[i][j].
If grid[i][j] equals grid[i-1][j], grid[i][j-1], and grid[i-1][j-1], then dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]); otherwise dp[i][j] = 1.
Iterate over the grid, fill the DP table, and accumulate the sum of all dp values. This sum is the total count of monochromatic squares.
Time complexity is O(m*n) and space complexity is O(m*n), but can be reduced to O(n) by keeping only the previous row.
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