52. N-Queens II (Hard)

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

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 the number of distinct solutions to the n-queens puzzle.

Example:

Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

Solutions

class Solution {

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

    int ans = 0;

    public int totalNQueens(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) {
            ans++;

            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;
    }
}

Incorrect Solutions

References

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