51. N-Queens

Problem Statement:

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 all distinct solutions to the n-queens puzzle. You may return the answer in any order.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space, respectively.

Example 1:

Input: n = 4
Output: [[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]

Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above

Example 2:

Input: n = 1
Output: [["Q"]]

Constraints:

1 <= n <= 9

Solution:

java

public class NQueens2 {

    public static List<List<String>> nQueens(int n) {
        // List of Lists of boards
        List<List<String>> allBoards = new ArrayList<>();
        char[][] board = new char[n][n];
        helper(board, allBoards, 0);
        return allBoards;
    }

    static void helper(char[][] board, List<List<String>> allBoards, int col) {
        // save board to allBoards after placing Queens on all possible cols
        if (col == board.length) {
            saveBoard(board, allBoards);
            return;
        }
        for (int row = 0; row < board.length; row++) {
            // if it's safe, place queen at row
            if (isSafe(row, col, board)) {
                board[row][col] = 'Q';
                helper(board, allBoards, col + 1);
                // backtracking
                board[row][col] = '.';
            }
        }
    }

    // Fn to check if it's safe to place Queen
    static boolean isSafe(int row, int col, char[][] board) {
        int len = board.length;
        // traverse in all cols to check if a queen is already present or not
        for (int i = 0; i < len; i++) {
            if (board[row][i] == 'Q')
                return false;
        }
        // traverse in all rows to check if a queen is already present or not
        for (int i = 0; i < len; i++) {
            if (board[i][col] == 'Q')
                return false;
        }
        // traverse through upper left diagonal to check if queen is present
        int r = row;
        for (int c = col; c >= 0 && r >= 0; r--, c--) {
            if (board[r][c] == 'Q')
                return false;
        }
        // traverse through upper right diagonal to check if queen is present
        r = row;
        for (int c = col; c < len && r >= 0; r--, c++) {
            if (board[r][c] == 'Q')
                return false;
        }
        // traverse through lower left diagonal to check if queen is present
        r = row;
        for (int c = col; c >= 0 && r < len; r++, c--) {
            if (board[r][c] == 'Q')
                return false;
        }
        // traverse through lower right diagonal to check if queen is present
        r = row;
        for (int c = col; c < len && r < len; r++, c++) {
            if (board[r][c] == 'Q')
                return false;
        }
        return true;
    }

    // Fn to save a board to List of Boards
    static void saveBoard(char[][] board, List<List<String>> allBoards) {
        int len = board.length;
        String row = "";
        List<String> newBoard = new ArrayList<>();
        for (int i = 0; i < len; i++) {
            row = "";
            for (int j = 0; j < len; j++) {
                if (board[i][j] == 'Q')
                    row += 'Q';
                else
                    row += '.';
            }
            // this adds the row to the newBoards -> "Q..." or "..Q."
            newBoard.add(row);
        }
        // this add the board to the list of boards -> [[..Q., Q..., ...Q, .Q..],...]
        allBoards.add(newBoard);
    }
}

51. N-Queens #

Problem Statement: #

Example 1: #

Example 2: #

Constraints: #

Solution: #

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