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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:
![N-Queens LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct] 2 queens](https://assets.leetcode.com/uploads/2020/11/13/queens.jpg)
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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