N Queens Problem in DSA Java - Time & Space Complexity

Time Complexity: N Queens Problem

O(n!)

Understanding Time Complexity

We want to understand how the time needed to solve the N Queens problem grows as the board size increases.

Specifically, how does the number of steps change when we try to place queens on larger boards?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class NQueens {
    private boolean isSafe(int row, int col, int[] positions) {
        for (int i = 0; i < row; i++) {
            if (positions[i] == col || Math.abs(positions[i] - col) == row - i) {
                return false;
            }
        }
        return true;
    }

    public int solve(int n, int row, int[] positions) {
        if (row == n) return 1;
        int count = 0;
        for (int col = 0; col < n; col++) {
            if (isSafe(row, col, positions)) {
                positions[row] = col;
                count += solve(n, row + 1, positions);
            }
        }
        return count;
    }
}

This code tries to place queens row by row on an n x n board, checking safety before placing each queen.

Identify Repeating Operations

• Primary operation: The recursive calls that try placing a queen in each column of the current row.
• How many times: For each row, it tries up to n columns, and for each safe placement, it recurses to the next row.
• Safety check: For each placement, it checks all previously placed queens (up to current row) to ensure no conflicts.

How Execution Grows With Input

As the board size n grows, the number of ways to place queens grows very fast because each row tries many columns and recurses deeply.

Input Size (n)	Approx. Operations
4	Hundreds of checks and recursive calls
8	Thousands to tens of thousands of checks
12	Millions of checks and recursive calls

Pattern observation: The operations grow faster than just multiplying by n each time; it grows roughly exponentially with n.

Final Time Complexity

Time Complexity: O(n!)

This means the time needed grows roughly like the factorial of n, which is very fast growth as n increases.

Common Mistake

[X] Wrong: "The time complexity is just O(n^2) because we have two loops."

[OK] Correct: The recursion creates many branches, not just two loops. Each row tries many columns, and this multiplies the work exponentially, not just squared.

Interview Connect

Understanding this complexity helps you explain why some problems need smart pruning or heuristics to solve efficiently, a skill valued in problem solving.

Self-Check

"What if we added memoization to remember safe placements? How would the time complexity change?"

Practice

1. What is the main goal of the N Queens Problem?

easy

A. Place queens only on the diagonal cells of the board

B. Place N queens on an NxN chessboard so that no two queens attack each other

C. Count the total number of queens on the board

D. Find the shortest path for a queen to move across the board

Solution

1. Step 1: Understand the problem statement

   The N Queens Problem requires placing N queens on an NxN chessboard so that no two queens threaten each other.

2. Step 2: Identify the correct goal

   This means no two queens share the same row, column, or diagonal.

3. Final Answer:

   Place N queens on an NxN chessboard so that no two queens attack each other -> Option B

4. Quick Check:

   Goal = Place queens safely [OK]

Hint: Remember: no two queens can share row, column, or diagonal [OK]

Common Mistakes:

• Thinking queens can share the same row or column
• Confusing the problem with counting queens
• Assuming queens only attack diagonally

2. Which of the following Java method signatures is correct for a recursive function solving the N Queens Problem?

easy

A. static int solveNQueens(int[][] board, int n, int row)

B. public void solveNQueens(int n, int row, int[][] board)

C. public static boolean solveNQueens(int row, int n, int[][] board)

D. boolean solveNQueens(int n, int[][] board, int row)

Solution

1. Step 1: Check method signature correctness

   The recursive method should return a boolean indicating success/failure, take current row, board size, and board state.

2. Step 2: Validate parameter order and modifiers

   public static boolean solveNQueens(int row, int n, int[][] board) matches this: public static boolean solveNQueens(int row, int n, int[][] board)

3. Final Answer:

   public static boolean solveNQueens(int row, int n, int[][] board) -> Option C

4. Quick Check:

   Correct signature = public static boolean solveNQueens(int row, int n, int[][] board) [OK]

Hint: Recursive method returns boolean and takes current row first [OK]

Common Mistakes:

• Wrong return type (e.g., void instead of boolean)
• Incorrect parameter order
• Missing static keyword if required

3. What will be the output of the following Java code snippet for N=4?

public class NQueensTest {
    static int count = 0;
    public static void main(String[] args) {
        int n = 4;
        int[][] board = new int[n][n];
        solveNQueens(0, n, board);
        System.out.println(count);
    }
    static boolean solveNQueens(int row, int n, int[][] board) {
        if (row == n) {
            count++;
            return true;
        }
        boolean res = false;
        for (int col = 0; col < n; col++) {
            if (isSafe(row, col, board, n)) {
                board[row][col] = 1;
                res = solveNQueens(row + 1, n, board) || res;
                board[row][col] = 0;
            }
        }
        return res;
    }
    static boolean isSafe(int row, int col, int[][] board, int n) {
        for (int i = 0; i < row; i++) {
            if (board[i][col] == 1) return false;
        }
        for (int i = row, j = col; i >= 0 && j >= 0; i--, j--) {
            if (board[i][j] == 1) return false;
        }
        for (int i = row, j = col; i >= 0 && j < n; i--, j++) {
            if (board[i][j] == 1) return false;
        }
        return true;
    }
}

medium

A. 2

B. 4

C. 1

D. 0

Solution

1. Step 1: Understand the code logic

   The code counts all valid solutions for 4 queens on a 4x4 board using backtracking.

2. Step 2: Recall known result for 4 queens

   It is a well-known fact that there are 2 valid solutions for the 4 queens problem.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Number of solutions for N=4 = 2 [OK]

Hint: Remember: 4 queens have exactly 2 solutions [OK]

Common Mistakes:

• Assuming only 1 solution exists
• Confusing number of solutions with board size
• Ignoring backtracking resets

4. Identify the bug in this snippet of the N Queens solver:

static boolean isSafe(int row, int col, int[][] board, int n) {
    for (int i = 0; i < row; i++) {
        if (board[i][col] == 1) return false;
    }
    for (int i = row, j = col; i >= 0 && j >= 0; i--, j--) {
        if (board[i][j] == 1) return false;
    }
    for (int i = row, j = col; i >= 0 && j <= n; i--, j++) {
        if (board[i][j] == 1) return false;
    }
    return true;
}

medium

A. It incorrectly checks columns beyond the board size

B. It does not check the left diagonal upwards

C. It does not check the current row for other queens

D. It misses checking the lower diagonals for threats

Solution

1. Step 1: Analyze the safety checks

   The column check and upper-left diagonal are correct.

2. Step 2: Identify the bug in upper-right diagonal

   The loop condition 'i >= 0 && j <= n' allows j to reach n (e.g., when col = n-1, after first iteration j = n), causing ArrayIndexOutOfBoundsException on board[i][n].

3. Final Answer:

   It incorrectly checks columns beyond the board size -> Option A

4. Quick Check:

   Upper-right diagonal: j <= n causes out-of-bounds [OK]

Hint: Always verify loop bounds in diagonal checks to avoid index errors [OK]

Common Mistakes:

• Forgetting off-by-one errors in diagonal loop conditions
• Overlooking array bounds in upper-right diagonal traversal
• Confusing upper vs lower diagonal checks

5. You want to modify the N Queens solver to print all distinct solutions for N=4. Which approach is best to achieve this?

hard

A. Stop recursion as soon as the first solution is found to save time

B. Place all queens in the first row and check if they attack each other

C. Use a greedy algorithm to place queens in the first available safe spot in each row

D. Modify the recursive function to store each valid board configuration in a list instead of returning after first success

Solution

1. Step 1: Understand the requirement

   We want to find and print all distinct solutions, not just one.

2. Step 2: Choose the correct approach

   Modifying the recursive function to continue searching after finding a solution and storing each valid board is the correct approach.

3. Step 3: Eliminate wrong options

   Stopping after first solution won't find all solutions. Greedy algorithm to place queens in the first available safe spot in each row and placing all queens in the first row and check if they attack each other are incorrect strategies.

4. Final Answer:

   Modify the recursive function to store each valid board configuration in a list instead of returning after first success -> Option D

5. Quick Check:

   Store all solutions by continuing recursion [OK]

Hint: Keep searching after each solution to collect all boards [OK]

Common Mistakes:

• Stopping recursion after first solution
• Using greedy instead of backtracking
• Placing multiple queens in one row