Approach Summary
Place queens row by row. Track used columns, diagonals (/), and anti-diagonals (\). Backtrack on conflicts.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n^2) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(n!) | O(n^2) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n^2)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
private boolean inbound(int i, int j, int n) {
return i >= 0 && j >= 0 && i < n && j < n;
}
private boolean isValid(List<String> board, int row, int col, int n) {
for (int i = 0; i < n; i++) {
if (board.get(i).charAt(col) == 'Q' || board.get(row).charAt(i) == 'Q') {
return false;
}
}
for (int i = 0; i < n; i++) {
if (inbound(row - i, col - i, n) && board.get(row - i).charAt(col - i) == 'Q') {
return false;
}
if (inbound(row + i, col + i, n) && board.get(row + i).charAt(col + i) == 'Q') {
return false;
}
if (inbound(row + i, col - i, n) && board.get(row + i).charAt(col - i) == 'Q') {
return false;
}
if (inbound(row - i, col + i, n) && board.get(row - i).charAt(col + i) == 'Q') {
return false;
}
}
return true;
}
private void solveNQueensUtil(List<List<String>> result, List<String> board, int row, int n) {
if (row == n) {
result.add(new ArrayList<>(board));
return;
}
for (int col = 0; col < n; col++) {
if (isValid(board, row, col, n)) {
StringBuilder sb = new StringBuilder(board.get(row));
sb.setCharAt(col, 'Q');
board.set(row, sb.toString());
solveNQueensUtil(result, board, row + 1, n);
sb.setCharAt(col, '.');
board.set(row, sb.toString());
}
}
}
public List<List<String>> solveNQueens(int n) {
List<List<String>> result = new ArrayList<>();
List<String> board = new ArrayList<>();
for (int i = 0; i < n; i++) {
StringBuilder sb = new StringBuilder();
for (int j = 0; j < n; j++) {
sb.append('.');
}
board.add(sb.toString());
}
solveNQueensUtil(result, board, 0, n);
return result;
}
}class Solution {
public:
bool inbound(int i, int j, int n){
if(i<0 || j<0 || i>=n || j>=n){
return false;
}
return true;
}
bool isValid(vector<string> &board, int row, int col, int n){
for(int i=0; i<n; i++){
if(board[i][col] == 'Q' || board[row][i] == 'Q'){
return false;
}
}
for(int i=0; i<n; i++){
if(inbound(row-i, col-i, n) && board[row-i][col-i] == 'Q'){
return false;
}
if(inbound(row+i, col+i, n) && board[row+i][col+i] == 'Q'){
return false;
}
if(inbound(row+i, col-i, n) && board[row+i][col-i] == 'Q'){
return false;
}
if(inbound(row-i, col+i, n) && board[row-i][col+i] == 'Q'){
return false;
}
}
return true;
}
void solveNQueensUtil(vector<vector<string>> &result, vector<string> &board, int row, int n){
if(row == n){
result.push_back(board);
return;
}
for(int col=0; col<n; col++){
if(isValid(board, row, col, n)){
board[row][col] = 'Q';
solveNQueensUtil(result, board, row+1, n);
board[row][col] = '.';
}
}
}
vector<vector<string>> solveNQueens(int n) {
vector<vector<string>> result;
vector<string> board(n, string(n, '.'));
solveNQueensUtil(result, board, 0, n);
return result;
}
};class Solution:
def n_queens(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for N Queens
*/
function nQueens(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(n!)
Space: O(n^2)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
private boolean inbound(int i, int j, int n) {
return i >= 0 && j >= 0 && i < n && j < n;
}
private boolean isValid(List<String> board, int row, int col, int n) {
for (int i = 0; i < n; i++) {
if (board.get(i).charAt(col) == 'Q' || board.get(row).charAt(i) == 'Q') {
return false;
}
}
for (int i = 0; i < n; i++) {
if (inbound(row - i, col - i, n) && board.get(row - i).charAt(col - i) == 'Q') {
return false;
}
if (inbound(row + i, col + i, n) && board.get(row + i).charAt(col + i) == 'Q') {
return false;
}
if (inbound(row + i, col - i, n) && board.get(row + i).charAt(col - i) == 'Q') {
return false;
}
if (inbound(row - i, col + i, n) && board.get(row - i).charAt(col + i) == 'Q') {
return false;
}
}
return true;
}
private void solveNQueensUtil(List<List<String>> result, List<String> board, int row, int n) {
if (row == n) {
result.add(new ArrayList<>(board));
return;
}
for (int col = 0; col < n; col++) {
if (isValid(board, row, col, n)) {
StringBuilder sb = new StringBuilder(board.get(row));
sb.setCharAt(col, 'Q');
board.set(row, sb.toString());
solveNQueensUtil(result, board, row + 1, n);
sb.setCharAt(col, '.');
board.set(row, sb.toString());
}
}
}
public List<List<String>> solveNQueens(int n) {
List<List<String>> result = new ArrayList<>();
List<String> board = new ArrayList<>();
for (int i = 0; i < n; i++) {
StringBuilder sb = new StringBuilder();
for (int j = 0; j < n; j++) {
sb.append('.');
}
board.add(sb.toString());
}
solveNQueensUtil(result, board, 0, n);
return result;
}
}class Solution {
public:
bool inbound(int i, int j, int n){
if(i<0 || j<0 || i>=n || j>=n){
return false;
}
return true;
}
bool isValid(vector<string> &board, int row, int col, int n){
for(int i=0; i<n; i++){
if(board[i][col] == 'Q' || board[row][i] == 'Q'){
return false;
}
}
for(int i=0; i<n; i++){
if(inbound(row-i, col-i, n) && board[row-i][col-i] == 'Q'){
return false;
}
if(inbound(row+i, col+i, n) && board[row+i][col+i] == 'Q'){
return false;
}
if(inbound(row+i, col-i, n) && board[row+i][col-i] == 'Q'){
return false;
}
if(inbound(row-i, col+i, n) && board[row-i][col+i] == 'Q'){
return false;
}
}
return true;
}
void solveNQueensUtil(vector<vector<string>> &result, vector<string> &board, int row, int n){
if(row == n){
result.push_back(board);
return;
}
for(int col=0; col<n; col++){
if(isValid(board, row, col, n)){
board[row][col] = 'Q';
solveNQueensUtil(result, board, row+1, n);
board[row][col] = '.';
}
}
}
vector<vector<string>> solveNQueens(int n) {
vector<vector<string>> result;
vector<string> board(n, string(n, '.'));
solveNQueensUtil(result, board, 0, n);
return result;
}
};class Solution:
def n_queens(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for N Queens
*/
function nQueens(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Backtracking Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Backtracking pattern.
Framework: Backtracking • Interview Prep: Tested at FAANG •
Explore all Backtracking problems & invariants →
How to Recognize This Pattern
- Place one per row; conflict check with col and two diagonal sets
- Classic constraint-satisfaction backtracking
Complexity Analysis
Time Complexity
O(n!)
Space Complexity
O(n)
Tags
Array Backtracking
Frequently Asked Questions
What is the optimal time and space complexity for N-Queens?
The optimal solution for N-Queens achieves O(n!) runtime complexity and requires O(n) auxiliary space. Time Complexity is O(n!) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(n^2) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that N-Queens requires the Backtracking pattern?
In technical interviews, look for key problem signals: Place one per row; conflict check with col and two diagonal sets; Classic constraint-satisfaction backtracking. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Backtracking strategy.
What critical edge cases must be handled in N-Queens?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask N-Queens?
N-Queens is a Hard-difficulty question tagged with medium recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Microsoft.