N-Queens LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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In this post, you will find the solution for the N-Queens in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – N-Queens– LeetCode Problem

N-Queens– LeetCode Problem

Problem:

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

N-Queens– LeetCode Solutions

N-Queens  in C++:

struct T {
  int left;   // sum of the subarray w/ max sum (starting from the first num)
  int right;  // sum of the subarray w/ max sum (ending at the the last num)
  int mid;    // sum of the subarray w/ max sum
  int sum;    // sum of the whole array
};

class Solution {
 public:
  int maxSubArray(vector<int>& nums) {
    const T t = divideAndConquer(nums, 0, nums.size() - 1);
    return t.mid;
  }

 private:
  T divideAndConquer(const vector<int>& nums, int l, int r) {
    if (l == r)
      return {nums[l], nums[l], nums[l], nums[l]};

    const int m = l + (r - l) / 2;
    const T t1 = divideAndConquer(nums, l, m);
    const T t2 = divideAndConquer(nums, m + 1, r);

    const int left = max(t1.left, t1.sum + t2.left);
    const int right = max(t1.right + t2.sum, t2.right);
    const int mid = max({t1.right + t2.left, t1.mid, t2.mid});
    const int sum = t1.sum + t2.sum;

    return {left, right, mid, sum};
  };
};

N-Queens in Java:

class Solution {
  public List<List<String>> solveNQueens(int n) {
    List<List<String>> ans = new ArrayList<>();
    char[][] board = new char[n][n];

    for (int i = 0; i < n; ++i)
      Arrays.fill(board[i], '.');

    dfs(n, 0, new boolean[n], new boolean[2 * n - 1], new boolean[2 * n - 1], board, ans);

    return ans;
  }

  private void dfs(int n, int i, boolean[] cols, boolean[] diag1, boolean[] diag2, char[][] board,
                   List<List<String>> ans) {
    if (i == n) {
      ans.add(construct(board));
      return;
    }

    for (int j = 0; j < cols.length; ++j) {
      if (cols[j] || diag1[i + j] || diag2[j - i + n - 1])
        continue;
      board[i][j] = 'Q';
      cols[j] = diag1[i + j] = diag2[j - i + n - 1] = true;
      dfs(n, i + 1, cols, diag1, diag2, board, ans);
      cols[j] = diag1[i + j] = diag2[j - i + n - 1] = false;
      board[i][j] = '.';
    }
  }

  private List<String> construct(char[][] board) {
    List<String> listBoard = new ArrayList<>();
    for (int i = 0; i < board.length; ++i)
      listBoard.add(String.valueOf(board[i]));
    return listBoard;
  }
}

N-Queens in Python:

class Solution:
  def solveNQueens(self, n: int) -> List[List[str]]:
    ans = []
    cols = [False] * n
    diag1 = [False] * (2 * n - 1)
    diag2 = [False] * (2 * n - 1)

    def dfs(i: int, board: List[int]) -> None:
      if i == n:
        ans.append(board)
        return

      for j in range(n):
        if cols[j] or diag1[i + j] or diag2[j - i + n - 1]:
          continue
        cols[j] = diag1[i + j] = diag2[j - i + n - 1] = True
        dfs(i + 1, board + ['.' * j + 'Q' + '.' * (n - j - 1)])
        cols[j] = diag1[i + j] = diag2[j - i + n - 1] = False

    dfs(0, [])

    return ans