51. N-Queens (Hard)

https://leetcode.com/problems/n-queens/

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

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:

Input: 4
Output: [
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.

Solutions

class Solution {

    private Map<String, Boolean> validityMap = new HashMap<>();
    private List<List<String>> ans = new ArrayList<>();

    public List<List<String>> solveNQueens(int n) {
        if (n < 1) {
            return ans;
        }

        int [] queens = new int[n];

        recurse(queens, 0);

        return ans;
    }

    private void recurse(int [] queens, int col) {
        if (col == queens.length) {
            List<String> result = convert(queens);

            ans.add(result);

            return;
        }

        for (int row = 0; row < queens.length; ++row) {
            for (int j = col + 1; j < queens.length; j++) {
                queens[j] = -1;
            }

            queens[col] = row;

            String key = Arrays.toString(queens);

            // We use validityMap to keep the track of valid and invalid sequence so that
            // we are able to reduce the computation overhead dramatically.
            if (!validityMap.containsKey(key)) {
                validityMap.put(key, isValid(queens, col, row));
            }

            // valid sequence, keep on digging into it.
            if (validityMap.get(key)) {
                recurse(queens, col + 1);
            }
        }
    }

    private boolean isValid(int [] queens, int col, int row) {

        // col is the last index, not the length

        for (int i = 0; i < col; ++i) {
            // previous point (i, queens[i])

            // horizontal comparision, check out if any pieces situated on same row
            if (queens[i] == row) {
                return false;
            }

            // Assume point A(a1, a2) and B(b1, b2) are diagonally arranged, abs(a1 - b1) == abs(a2 - b2).
            // We can imagine that with point A and B as the diagonal points, we can form a square.
            // As a square, all the edges should be of same length.

            // diagonal comparision
            if (Math.abs(col - i) == Math.abs(row - queens[i])) {
                return false;
            }
        }

        return true;
    }

    private List<String> convert(int [] queens) {
        int len = queens.length;

        List<String> ans = new ArrayList<>();
        for (int c = 0; c < len; c++) {
            String row = "";
            for (int r = 0; r < len; r++) {
                if (r == queens[c]) {
                    row += "Q";
                } else {
                    row += ".";
                }
            }

            ans.add(row);
        }

        return ans;
    }
}

Incorrect Solutions

References

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