What is the time complexity of N-Queens?

The optimal solution has a time complexity of O(N!).

What is the space complexity of N-Queens?

The space complexity is O(N^2).

What data structures are used to solve N-Queens?

This problem uses Array, Backtracking. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving N-Queens?

Insight 1: Backtracking is the most suitable approach as it explores different placement possibilities systematically and abandons branches that lead to invalid solutions. Insight 2: Using sets to track occupied columns and diagonals allows for efficient checking of whether a placement is valid (O(1) lookup).

Which companies ask LeetCode 51. N-Queens in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Citadel and 11 more companies.

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N-Queens - LeetCode 51 Solution

N-Queens - Complete Solution Guide

N-Queens is LeetCode problem 51, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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:

Detailed Explanation

The N-Queens puzzle asks us to place 'n' queens on an 'n x n' chessboard such that no two queens threaten each other. This means no two queens can be in the same row, column, or diagonal. The goal is to find all distinct arrangements (solutions) of placing the queens on the board. The input is a single integer 'n' representing the size of the board. The output is a list of lists of strings, where each inner list represents a solution. Each string in the inner list represents a row of the chessboard, with 'Q' denoting a queen and '.' denoting an empty space. The constraint is that 'n' will be between 1 and 9.

Solution Approach

The provided code utilizes a backtracking algorithm to systematically explore all possible queen placements. It maintains sets to track columns, positive diagonals, and negative diagonals that are already occupied by queens. The algorithm recursively attempts to place a queen in each column of each row. If a placement is valid (doesn't conflict with existing queens), it adds the queen to the board and recursively calls itself for the next row. If the current row is the last row, a valid solution has been found and added to the results. After exploring a branch of placements, the algorithm 'backtracks' by removing the queen from the current position to explore other possibilities.

Step-by-Step Algorithm

Step 1: Initialize an empty board (list of lists of strings) filled with '.' representing empty spaces.
Step 2: Initialize three sets: 'cols', 'pos_diagonals', and 'neg_diagonals' to track occupied columns and diagonals.
Step 3: Define a recursive 'backtrack' function that takes the current row 'r' as input.
Step 4: Base Case: If 'r' equals 'n', it means all 'n' queens have been placed successfully. Convert the current board configuration to a list of strings and append it to the 'results' list.
Step 5: Iterate through each column 'c' in the current row 'r'.
Step 6: Check if placing a queen at board[r][c] is valid. This is done by checking if 'c' is in 'cols', 'r - c' is in 'pos_diagonals', or 'r + c' is in 'neg_diagonals'. If any of these conditions are true, the placement is invalid, and we continue to the next column.
Step 7: If the placement is valid, add 'c' to 'cols', 'r - c' to 'pos_diagonals', and 'r + c' to 'neg_diagonals'. Also, place a queen 'Q' at board[r][c].
Step 8: Recursively call 'backtrack(r + 1)' to attempt to place a queen in the next row.
Step 9: After the recursive call returns, 'backtrack'. This means remove 'c' from 'cols', 'r - c' from 'pos_diagonals', and 'r + c' from 'neg_diagonals'. Also, remove the queen 'Q' from board[r][c] by setting it back to '.'. This allows exploring other possible placements in the current row.
Step 10: Start the backtracking process by calling 'backtrack(0)' to place the first queen in row 0.
Step 11: Return the 'results' list, which contains all valid N-Queens solutions.

Key Insights

Insight 1: Backtracking is the most suitable approach as it explores different placement possibilities systematically and abandons branches that lead to invalid solutions.
Insight 2: Using sets to track occupied columns and diagonals allows for efficient checking of whether a placement is valid (O(1) lookup).
Insight 3: The two types of diagonals can be represented by r - c (positive slope) and r + c (negative slope), which simplifies conflict detection.

Complexity Analysis

Time Complexity: O(N!)

Space Complexity: O(N^2)

Topics

This problem involves: Array, Backtracking.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Citadel, Goldman Sachs, IBM, Meta, Microsoft, Salesforce, Samsung, Snowflake, TikTok, Uber, Yahoo, Zoho.

N-Queens | LeetCode 51

Hard

Array Backtracking

Time: O(N!)Space: O(N^2)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> results = new ArrayList<>();
        Set<Integer> cols = new HashSet<>();
        Set<Integer> pos_diagonals = new HashSet<>(); // r - c
        Set<Integer> neg_diagonals = new HashSet<>(); // r + c

        char[][] board = new char[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] = '.';
            }
        }

        function backtrack(int r){
            if (r == n) {
                List<String> copy = new ArrayList<>();
                for (int i = 0; i < n; i++) {
                    copy.add(new String(board[i]));
                }
                results.add(copy);
                return;
            }

            for (int c = 0; c < n; c++) {
                if (cols.contains(c) || pos_diagonals.contains(r - c) || neg_diagonals.contains(r + c)) {
                    continue;
                }

                cols.add(c);
                pos_diagonals.add(r - c);
                neg_diagonals.add(r + c);
                board[r][c] = 'Q';

                backtrack(r + 1);

                cols.remove(c);
                pos_diagonals.remove(r - c);
                neg_diagonals.remove(r + c);
                board[r][c] = '.';
            }
        }

        backtrack(0);
        return results;
    }

    private void backtrack(int r, int n, List<List<String>> results, Set<Integer> cols, Set<Integer> pos_diagonals, Set<Integer> neg_diagonals, char[][] board) {
         if (r == n) {
                List<String> copy = new ArrayList<>();
                for (int i = 0; i < n; i++) {
                    copy.add(new String(board[i]));
                }
                results.add(copy);
                return;
            }

            for (int c = 0; c < n; c++) {
                if (cols.contains(c) || pos_diagonals.contains(r - c) || neg_diagonals.contains(r + c)) {
                    continue;
                }

                cols.add(c);
                pos_diagonals.add(r - c);
                neg_diagonals.add(r + c);
                board[r][c] = 'Q';

                backtrack(r + 1, n, results, cols, pos_diagonals, neg_diagonals, board);

                cols.remove(c);
                pos_diagonals.remove(r - c);
                neg_diagonals.remove(r + c);
                board[r][c] = '.';
            }
    }
}

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Problem Info

Description

Problem 51: N-Queens - Detailed explanation and solution approaches.

Topics

Array Backtracking

Companies

AdobeAmazonAppleBloombergCitadelGoldman SachsIBMMeta

Stats

Difficulty:Hard

Languages:4

Companies:16

Problem #:51

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize an empty board (list of lists of strings) filled with '.' representing empty spaces.
2
Step 2: Initialize three sets: 'cols', 'pos_diagonals', and 'neg_diagonals' to track occupied columns and diagonals.
3
Step 3: Define a recursive 'backtrack' function that takes the current row 'r' as input.
4
Step 4: Base Case: If 'r' equals 'n', it means all 'n' queens have been placed successfully. Convert the current board configuration to a list of strings and append it to the 'results' list.
5
Step 5: Iterate through each column 'c' in the current row 'r'.
6
Step 6: Check if placing a queen at board[r][c] is valid. This is done by checking if 'c' is in 'cols', 'r - c' is in 'pos_diagonals', or 'r + c' is in 'neg_diagonals'. If any of these conditions are true, the placement is invalid, and we continue to the next column.
7
Step 7: If the placement is valid, add 'c' to 'cols', 'r - c' to 'pos_diagonals', and 'r + c' to 'neg_diagonals'. Also, place a queen 'Q' at board[r][c].
8
Step 8: Recursively call 'backtrack(r + 1)' to attempt to place a queen in the next row.
9
Step 9: After the recursive call returns, 'backtrack'. This means remove 'c' from 'cols', 'r - c' from 'pos_diagonals', and 'r + c' from 'neg_diagonals'. Also, remove the queen 'Q' from board[r][c] by setting it back to '.'. This allows exploring other possible placements in the current row.
10
Step 10: Start the backtracking process by calling 'backtrack(0)' to place the first queen in row 0.
11
Step 11: Return the 'results' list, which contains all valid N-Queens solutions.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N!)

💾

Space Complexity

O(N^2)

📝

Why?

Time Complexity: In the worst-case scenario, we explore all possible permutations of queen placements. In the first row, we have n choices, then in the next row, potentially n choices again, and so on. This results in O(n!) time complexity. Although not all branches will be explored to the deepest level due to pruning invalid states, the upper bound remains O(n!). Space Complexity: O(N^2) primarily due to the board representation. The recursion depth is O(N), and the sets 'cols', 'pos_diagonals', and 'neg_diagonals' each store at most N elements, so these contribute O(N) space. However the board itself requires O(N^2) to store character data.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Bit Manipulation

Bit manipulation can be used to optimize the space complexity. Instead of using sets to track occupied columns and diagonals, we can use integers where each bit represents the availability of a column or diagonal. This can slightly improve the performance.

Time: O(N!)Space: O(N)

Trade-offs: Bit manipulation can be faster and more space-efficient, but it might be more complex to implement and understand compared to using sets.

✨

Key Insights

1Insight 1: Backtracking is the most suitable approach as it explores different placement possibilities systematically and abandons branches that lead to invalid solutions.
2Insight 2: Using sets to track occupied columns and diagonals allows for efficient checking of whether a placement is valid (O(1) lookup).
3Insight 3: The two types of diagonals can be represented by r - c (positive slope) and r + c (negative slope), which simplifies conflict detection.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: n = 1. The solution should return a single board with a queen in the top-left corner: [['Q']]. The provided code correctly handles this case.
Edge case 2: n = 2 or n = 3. There are no valid solutions. The backtracking algorithm will explore all possibilities and return an empty list. The provided code correctly handles this case.
Edge case 3: Input validation: While the constraints specify 1 <= n <= 9, a more robust solution would include a check at the beginning of the function to ensure that n is within the allowed range, returning an empty list or throwing an exception if it is not. The provided solutions do not have explicit input validation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not backtracking correctly. Failing to remove the queen from the board and the column/diagonal sets after exploring a branch can lead to incorrect solutions.

✗Common mistake 2: Incorrectly calculating diagonal indices. The diagonals need to be represented in a way that distinguishes them uniquely. Using just 'r - c' and 'r + c' is crucial. Forgetting to offset 'r - c' by 'n - 1' in the C code for the pos_diagonals array index leads to errors.

✗Common mistake 3: Not creating a deep copy of the board before adding it to the results. If the same board object is modified in each iteration, all solutions in the 'results' list will end up being the same.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Constraint satisfaction problems. The N-Queens problem is a classic example of a constraint satisfaction problem (CSP). Many real-world problems can be modeled as CSPs, such as scheduling, resource allocation, and circuit design.

🏦Real-world scenario 2: Game AI. Backtracking algorithms are often used in game AI to explore possible moves and determine the best strategy. For example, in a chess-playing AI, backtracking can be used to evaluate different move sequences and choose the one that leads to the highest score.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "N-Queens"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?