# N-Queens Problem

!!! question

    According to the rules of chess, a queen can attack any piece in the same row, column, or diagonal. Given $n$ queens and an $n \times n$ chessboard, find an arrangement such that no two queens can attack each other.

As shown in the figure below, when $n = 4$, there are two solutions that can be found. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, which provide all the choices `choices`. During the process of placing queens one by one, the chessboard state changes continuously, and the chessboard at each moment represents the state `state`.

![Solution to the 4-queens problem](n_queens_problem.assets/solution_4_queens.png)

The figure below illustrates the three constraints of this problem: **multiple queens cannot be in the same row, the same column, or on the same diagonal**. It is worth noting that diagonals are divided into two types: the main diagonal `\` and the anti-diagonal `/`.

![Constraints of the n-queens problem](n_queens_problem.assets/n_queens_constraints.png)

### Row-By-Row Placement Strategy

Since both the number of queens and the number of rows on the chessboard are $n$, we can easily derive a conclusion: **each row of the chessboard allows one and only one queen to be placed**.

This means we can adopt a row-by-row placement strategy: starting from the first row, place one queen in each row until the last row is completed.

The figure below shows the row-by-row placement process for the 4-queens problem. Due to space limitations, the figure only expands one search branch of the first row, and all schemes that violate the column or diagonal constraints are pruned.

![Row-by-row placement strategy](n_queens_problem.assets/n_queens_placing.png)

Essentially, **the row-by-row placement strategy serves a pruning function**, as it avoids all search branches where multiple queens appear in the same row.

### Column and Diagonal Pruning

To satisfy the column constraint, we can use a boolean array `cols` of length $n$ to record whether each column has a queen. Before each placement decision, we use `cols` to prune columns that already have queens, and dynamically update the state of `cols` during backtracking.

!!! tip

    Please note that the origin of the matrix is located in the upper-left corner, where the row index increases from top to bottom, and the column index increases from left to right.

So how do we handle diagonal constraints? Consider a square on the chessboard with row and column indices $(row, col)$. If we select a specific main diagonal in the matrix, we find that all squares on that diagonal have the same difference between their row and column indices, **meaning that $row - col$ is a constant value for all squares on the main diagonal**.

In other words, if two squares satisfy $row_1 - col_1 = row_2 - col_2$, they must be on the same main diagonal. Using this pattern, we can use the array `diags1` shown in the figure below to record whether there is a queen on each main diagonal.

Similarly, **for all squares on an anti-diagonal, the sum $row + col$ is a constant value**. We can likewise use the array `diags2` to handle anti-diagonal constraints.

![Handling column and diagonal constraints](n_queens_problem.assets/n_queens_cols_diagonals.png)

### Code Implementation

Please note that in an $n \times n$ square matrix, the range of $row - col$ is $[-n + 1, n - 1]$, and the range of $row + col$ is $[0, 2n - 2]$. Therefore, the number of both main diagonals and anti-diagonals is $2n - 1$, meaning the length of both arrays `diags1` and `diags2` is $2n - 1$.

```src
[file]{n_queens}-[class]{}-[func]{n_queens}
```

Placing $n$ queens row by row, considering the column constraint, from the first row to the last row there are $n$, $n-1$, $\dots$, $2$, $1$ choices, using $O(n!)$ time. When recording a solution, it is necessary to copy the matrix `state` and add it to `res`, and the copy operation uses $O(n^2)$ time. Therefore, **the overall time complexity is $O(n! \cdot n^2)$**. In practice, pruning based on diagonal constraints can also significantly reduce the search space, so the search efficiency is often better than the time complexity mentioned above.

The array `state` uses $O(n^2)$ space, and the arrays `cols`, `diags1`, and `diags2` each use $O(n)$ space. The maximum recursion depth is $n$, using $O(n)$ stack frame space. Therefore, **the space complexity is $O(n^2)$**.