g0001_0100.s0052_n_queens_ii.Solution Maven / Gradle / Ivy

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package g0001_0100.s0052_n_queens_ii;

// #Hard #Backtracking #2022_06_17_Time_1_ms_(97.55%)_Space_41.4_MB_(37.24%)

/**
 * 52 - N-Queens II\.
 *
 * Hard
 *
 * The **n-queens** puzzle is the problem of placing `n` queens on an `n x n` chessboard such that no two queens attack each other.
 *
 * Given an integer `n`, return _the number of distinct solutions to the **n-queens puzzle**_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/11/13/queens.jpg)
 *
 * **Input:** n = 4
 *
 * **Output:** 2
 *
 * **Explanation:** There are two distinct solutions to the 4-queens puzzle as shown. 
 *
 * **Example 2:**
 *
 * **Input:** n = 1
 *
 * **Output:** 1 
 *
 * **Constraints:**
 *
 * *   `1 <= n <= 9`
**/
public class Solution {
    public int totalNQueens(int n) {
        boolean[] row = new boolean[n];
        boolean[] col = new boolean[n];
        boolean[] diagonal = new boolean[n + n - 1];
        boolean[] antiDiagonal = new boolean[n + n - 1];
        return totalNQueens(n, 0, row, col, diagonal, antiDiagonal);
    }

    private int totalNQueens(
            int n,
            int r,
            boolean[] row,
            boolean[] col,
            boolean[] diagonal,
            boolean[] antiDiagonal) {
        if (r == n) {
            return 1;
        }
        int count = 0;
        for (int c = 0; c < n; c++) {
            if (!row[r] && !col[c] && !diagonal[r + c] && !antiDiagonal[r - c + n - 1]) {
                row[r] = col[c] = diagonal[r + c] = antiDiagonal[r - c + n - 1] = true;
                count += totalNQueens(n, r + 1, row, col, diagonal, antiDiagonal);
                row[r] = col[c] = diagonal[r + c] = antiDiagonal[r - c + n - 1] = false;
            }
        }
        return count;
    }
}