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

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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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LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

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  • Mathematics/Basic Logical Based Questions
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  • Database
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  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemN-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:

queens
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

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