The single-pass constraint is what makes this interesting.
Clarify that 'valid' means no duplicates among filled cells, then propose a single-pass solution using hash sets (or boolean arrays) for rows, columns, and boxes. Iterate over each cell once, compute the box index, and check for duplicates before inserting.
Pro tip: Emphasize that the board is partially filled, so you only need to validate existing numbers; also mention that you can use bitmasks for O(1) space per row/column/box, which is a strong optimization.
Confirm that the board is 9x9, digits are '1'-'9', '.' means empty, and validity only concerns duplicates among filled cells. Ask if the board is guaranteed to be 9x9 and if input is mutable.
Use three sets (or boolean arrays) per row, column, and 3x3 sub-box to track seen digits. Alternatively, use bitmasks (integers) for constant space and faster checks.
Loop through each cell (r, c). If the cell is not '.', compute the box index as (r/3)*3 + c/3. Check if the digit already exists in the corresponding row, column, or box set; if yes, return false. Otherwise, add the digit to all three sets.
After the loop, return true. State that time complexity is O(1) because the board size is fixed (81 cells), but generally O(n^2) for an n x n board. Space complexity is O(1) with fixed-size sets or bitmasks.
Mention that using bitmasks reduces space and may improve speed. Also note that early termination on duplicate detection saves time. If asked, explain how to extend to full Sudoku solver.
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