← Bytedance Interview Insights
Knew it was backtracking pretty quick but the implementation took me longer than I'd like to admit.
Use backtracking to place queens row by row, maintaining sets for columns and diagonals to ensure O(1) conflict checks. After finding a valid placement, convert the board state to the required list of strings format. Then analyze the time complexity as O(n!) and space complexity as O(n^2) for the output plus O(n) for auxiliary data structures.
Pro tip: Mention that the number of solutions grows factorially, so for large n the output size dominates; you can optimize by using bitmasks for conflict detection to reduce constant factors. Also, clarify that the space complexity includes the output, which is O(n^2 * number of solutions).
Confirm that n is a positive integer, and that each solution is a list of n strings, each of length n, with 'Q' and '.' representing queens and empty spaces. Discuss edge cases like n=1 (one solution) and n=2,3 (no solutions).
Use a recursive function that places a queen in each row, trying all columns. Maintain sets for columns, diagonals (row+col), and anti-diagonals (row-col) to check validity in O(1).
When a valid placement for all n rows is found, convert the current board state (e.g., an array of column indices) into the list of strings format and add to results. Backtrack by removing the queen and updating sets.
Time: O(n!) because in the worst case we explore all permutations, though pruning reduces it. Space: O(n^2 * S) for the output where S is the number of solutions, plus O(n) for recursion and sets.
Mention using bitmasks for faster conflict checks, or symmetry reduction to halve the search space. Compare iterative vs recursive approaches and their memory implications.
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