LeetCode #51 — HARD

N-Queens

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Backtracking

Example 1

Example 2

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

import java.util.*;

class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> ans = new ArrayList<>();
        char[][] board = new char[n][n];
        for (int r = 0; r < n; r++) Arrays.fill(board[r], '.');

        // Columns, main diagonals, anti-diagonals currently occupied by queens.
        boolean[] cols = new boolean[n];
        boolean[] diag1 = new boolean[2 * n]; // row - col + n
        boolean[] diag2 = new boolean[2 * n]; // row + col

        backtrack(0, n, board, cols, diag1, diag2, ans);
        return ans;
    }

    private void backtrack(int row, int n, char[][] board, boolean[] cols, boolean[] diag1,
                           boolean[] diag2, List<List<String>> ans) {
        if (row == n) {
            List<String> built = new ArrayList<>();
            for (char[] r : board) built.add(new String(r));
            ans.add(built);
            return;
        }

        for (int col = 0; col < n; col++) {
            int d1 = row - col + n;
            int d2 = row + col;
            if (cols[col] || diag1[d1] || diag2[d2]) continue;

            cols[col] = diag1[d1] = diag2[d2] = true;
            board[row][col] = 'Q';
            backtrack(row + 1, n, board, cols, diag1, diag2, ans);
            board[row][col] = '.';
            cols[col] = diag1[d1] = diag2[d2] = false;
        }
    }
}

import "strings"

func solveNQueens(n int) [][]string {
    res := [][]string{}
    board := make([][]byte, n)
    for i := 0; i < n; i++ {
        board[i] = []byte(strings.Repeat(".", n))
    }

    cols := make([]bool, n)
    diag1 := make([]bool, 2*n) // row - col + n
    diag2 := make([]bool, 2*n) // row + col

    var backtrack func(row int)
    backtrack = func(row int) {
        if row == n {
            built := make([]string, n)
            for i := 0; i < n; i++ {
                built[i] = string(board[i])
            }
            res = append(res, built)
            return
        }

        for col := 0; col < n; col++ {
            d1 := row - col + n
            d2 := row + col
            if cols[col] || diag1[d1] || diag2[d2] {
                continue
            }

            cols[col], diag1[d1], diag2[d2] = true, true, true
            board[row][col] = 'Q'
            backtrack(row + 1)
            board[row][col] = '.'
            cols[col], diag1[d1], diag2[d2] = false, false, false
        }
    }

    backtrack(0)
    return res
}

from typing import List

class Solution:
    def solveNQueens(self, n: int) -> List[List[str]]:
        ans: List[List[str]] = []
        board = [['.'] * n for _ in range(n)]
        cols = set()
        diag1 = set()  # row - col
        diag2 = set()  # row + col

        def backtrack(row: int) -> None:
            if row == n:
                ans.append([''.join(r) for r in board])
                return

            for col in range(n):
                d1 = row - col
                d2 = row + col
                if col in cols or d1 in diag1 or d2 in diag2:
                    continue

                cols.add(col)
                diag1.add(d1)
                diag2.add(d2)
                board[row][col] = 'Q'

                backtrack(row + 1)

                board[row][col] = '.'
                cols.remove(col)
                diag1.remove(d1)
                diag2.remove(d2)

        backtrack(0)
        return ans

impl Solution {
    pub fn solve_n_queens(n: i32) -> Vec<Vec<String>> {
        let n = n as usize;
        let mut board = vec![vec!['.'; n]; n];
        let mut cols = vec![false; n];
        let mut diag1 = vec![false; 2 * n]; // row - col + n
        let mut diag2 = vec![false; 2 * n]; // row + col
        let mut ans = Vec::new();

        fn backtrack(
            row: usize,
            n: usize,
            board: &mut Vec<Vec<char>>,
            cols: &mut Vec<bool>,
            diag1: &mut Vec<bool>,
            diag2: &mut Vec<bool>,
            ans: &mut Vec<Vec<String>>,
        ) {
            if row == n {
                ans.push(board.iter().map(|r| r.iter().collect()).collect());
                return;
            }

            for col in 0..n {
                let d1 = row + n - col;
                let d2 = row + col;
                if cols[col] || diag1[d1] || diag2[d2] {
                    continue;
                }

                cols[col] = true;
                diag1[d1] = true;
                diag2[d2] = true;
                board[row][col] = 'Q';

                backtrack(row + 1, n, board, cols, diag1, diag2, ans);

                board[row][col] = '.';
                cols[col] = false;
                diag1[d1] = false;
                diag2[d2] = false;
            }
        }

        backtrack(0, n, &mut board, &mut cols, &mut diag1, &mut diag2, &mut ans);
        ans
    }
}

function solveNQueens(n: number): string[][] {
  const ans: string[][] = [];
  const board: string[][] = Array.from({ length: n }, () => Array(n).fill('.'));
  const cols = new Set<number>();
  const diag1 = new Set<number>(); // row - col
  const diag2 = new Set<number>(); // row + col

  const backtrack = (row: number): void => {
    if (row === n) {
      ans.push(board.map((r) => r.join('')));
      return;
    }

    for (let col = 0; col < n; col++) {
      const d1 = row - col;
      const d2 = row + col;
      if (cols.has(col) || diag1.has(d1) || diag2.has(d2)) continue;

      cols.add(col);
      diag1.add(d1);
      diag2.add(d2);
      board[row][col] = 'Q';

      backtrack(row + 1);

      board[row][col] = '.';
      cols.delete(col);
      diag1.delete(d1);
      diag2.delete(d2);
    }
  };

  backtrack(0);
  return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2 × n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.