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N-Queens

Hard

Adobe

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Topics:

RecursionArrays

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

N-Queens

Hard

Adobe

0 views

Topics:

RecursionArrays

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

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the value range of `n`? Is there a maximum or minimum size I should consider?
If no solutions exist, what should the function return? Should it be an empty list or something else?
Do you want all possible solutions, or is it sufficient to return any one valid arrangement?
How should the board be represented in the output? Should it be a list of strings, a 2D array, or another data structure?
Is the order of the solutions in the output important?

Brute Force Solution

Approach

The N-Queens problem asks us to place queens on a chessboard so that none can attack each other. The brute force method is like trying every single possible arrangement of queens on the board until you find one that works.

Here's how the algorithm would work step-by-step:

Start by placing a queen in the first possible spot on the board.
Then, try placing a second queen in every possible spot that isn't already under attack by the first queen.
Continue placing queens one by one, each time considering every available spot and checking if it's safe from all the previously placed queens.
If you reach a point where you can't place a queen because every spot is under attack, go back and try a different position for the previous queen you placed.
If you successfully place all the queens on the board without any of them attacking each other, you've found a solution.
Keep trying different arrangements until you find all the possible solutions.

Code Implementation

def solve_n_queens_brute_force(number_of_queens):
    solutions = []
    
    def is_safe(board_state, row_index, column_index):
        for previous_row_index in range(row_index):
            # Check same column
            if board_state[previous_row_index] == column_index:
                return False
            
            # Check diagonals
            column_distance = abs(board_state[previous_row_index] - column_index)
            row_distance = row_index - previous_row_index
            if column_distance == row_distance:
                return False
        return True
    
    def find_all_arrangements(row_index, board_state):
        # Base case: All queens are placed
        if row_index == number_of_queens:
            solutions.append(board_state[:])
            return
        
        for column_index in range(number_of_queens):
            # Check if this position is safe
            if is_safe(board_state, row_index, column_index):

                board_state[row_index] = column_index

                # Recursively try to place the next queen
                find_all_arrangements(row_index + 1, board_state)

    # Initialize the board state
    board_state = [0] * number_of_queens

    # Start the recursive process from the first row
    find_all_arrangements(0, board_state)
    
    return solutions

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n!) – The brute force N-Queens approach explores numerous possible queen placements. In the worst case, for each row, we might need to explore almost all columns. The first queen has n choices, the second potentially n-1 (if no attacks yet), the third n-2, and so on. This leads to roughly n * (n-1) * (n-2) * ... * 1, which is n factorial. Thus, the time complexity is O(n!).

Space Complexity

O(N) – The algorithm uses recursion to explore possible queen placements. The maximum depth of the recursion is N, as we place one queen per row. Each recursive call requires stack space to store the state (local variables, return address), so the maximum space used by the call stack is proportional to N. Therefore, the auxiliary space complexity is O(N).

Optimal Solution

Approach

The N-Queens problem involves placing queens on a chessboard so that no two queens threaten each other. The optimal solution strategically places queens row by row, avoiding the need to explore invalid placements. It leverages a key insight: once a queen is placed, we know the squares that are no longer available.

Here's how the algorithm would work step-by-step:

Start by placing a queen in the first available column of the first row.
Mark all the squares that this queen attacks: those in the same column, the same diagonal going up and to the right, and the same diagonal going down and to the right.
Move to the next row. Find the first column that is not under attack and place a queen there.
Again, mark all the squares that this new queen attacks.
Keep repeating this process for each row. If, at any row, there are no available columns that are safe, it means that the placement in the previous rows was incorrect. In that case, go back to the previous row and try placing the queen in a different column.
If you reach the last row and you are able to place a queen, then you have found a valid solution.
To find all possible solutions, continue to backtrack and try different column placements for the queens in earlier rows until all possibilities have been exhausted.

Code Implementation

def solve_n_queens(number_of_queens):
    solutions = []
    board_state = [0] * number_of_queens

    def is_safe(row_index, column_index):
        for previous_row_index in range(row_index):
            previous_column_index = board_state[previous_row_index]
            if previous_column_index == column_index:
                return False
            if abs(previous_column_index - column_index) == row_index - previous_row_index:
                return False
        return True

    def find_solutions(row_index):
        # Base case: All queens are placed.
        if row_index == number_of_queens:
            solutions.append(board_state[:])
            return

        # Try placing a queen in each column of the current row.
        for column_index in range(number_of_queens):
            if is_safe(row_index, column_index):

                # Place the queen if the position is safe
                board_state[row_index] = column_index

                # Recursively find solutions for the next row.
                find_solutions(row_index + 1)

                # Backtrack: Remove the queen and try the next column.

    find_solutions(0)
    return solutions

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n!) – The algorithm explores possible queen placements row by row. In the worst case, for the first row, we might try all n columns. For the second row, again potentially n columns, and so on. This leads to a decision tree where, at each level (row), we have up to n choices. Although backtracking helps prune the search space, in the worst-case scenario, the algorithm explores a significant portion of the possible permutations of queen placements. Therefore, the upper bound of the time complexity is O(n!).

Space Complexity

O(N) – The space complexity is primarily determined by the depth of the recursion stack, which can go as deep as N in the worst case (where N is the number of rows or columns on the chessboard). Additionally, the solution described might utilize auxiliary data structures to keep track of attacked squares (columns and diagonals). The space required for these is also related to N. Therefore, the auxiliary space used is O(N).

Edge Cases

Case	How to Handle
N is zero or negative	Return an empty list as there is no valid board configuration for non-positive N.
N is equal to 1	Return a list containing a single solution [['Q']] as a 1x1 board trivially allows placing one queen.
N is 2 or 3	Return an empty list, as no valid solution exists for N=2 or N=3 due to the queens being able to attack each other.
Large N values (e.g., N > 12)	The backtracking algorithm's time complexity increases exponentially, potentially leading to stack overflow or timeout errors; consider iterative solutions or pruning optimizations.
Integer overflow when calculating diagonals for very large N	Ensure data types used for diagonal calculations can handle large values, potentially using long or BigInteger if needed to prevent overflows.
Multiple valid solutions exist	The backtracking algorithm should explore all possible valid placements and collect all distinct solutions in a result list.
Memory exhaustion for large N due to storing all solutions	If memory is a constraint, consider alternative approaches like returning only the count of solutions, or generating solutions on demand without storing all of them simultaneously.
Floating point errors with approximations of diagonals.	Ensure the code relies on integer arithmetic when calculating the positions of the diagonals, rather than relying on floating point computations.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the value range of `n`? Is there a maximum or minimum size I should consider?
If no solutions exist, what should the function return? Should it be an empty list or something else?
Do you want all possible solutions, or is it sufficient to return any one valid arrangement?
How should the board be represented in the output? Should it be a list of strings, a 2D array, or another data structure?
Is the order of the solutions in the output important?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by placing a queen in the first possible spot on the board.
Then, try placing a second queen in every possible spot that isn't already under attack by the first queen.
Continue placing queens one by one, each time considering every available spot and checking if it's safe from all the previously placed queens.
If you reach a point where you can't place a queen because every spot is under attack, go back and try a different position for the previous queen you placed.
If you successfully place all the queens on the board without any of them attacking each other, you've found a solution.
Keep trying different arrangements until you find all the possible solutions.

Code Implementation

def solve_n_queens_brute_force(number_of_queens):
    solutions = []
    
    def is_safe(board_state, row_index, column_index):
        for previous_row_index in range(row_index):
            # Check same column
            if board_state[previous_row_index] == column_index:
                return False
            
            # Check diagonals
            column_distance = abs(board_state[previous_row_index] - column_index)
            row_distance = row_index - previous_row_index
            if column_distance == row_distance:
                return False
        return True
    
    def find_all_arrangements(row_index, board_state):
        # Base case: All queens are placed
        if row_index == number_of_queens:
            solutions.append(board_state[:])
            return
        
        for column_index in range(number_of_queens):
            # Check if this position is safe
            if is_safe(board_state, row_index, column_index):

                board_state[row_index] = column_index

                # Recursively try to place the next queen
                find_all_arrangements(row_index + 1, board_state)

    # Initialize the board state
    board_state = [0] * number_of_queens

    # Start the recursive process from the first row
    find_all_arrangements(0, board_state)
    
    return solutions

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start by placing a queen in the first available column of the first row.
Mark all the squares that this queen attacks: those in the same column, the same diagonal going up and to the right, and the same diagonal going down and to the right.
Move to the next row. Find the first column that is not under attack and place a queen there.
Again, mark all the squares that this new queen attacks.
Keep repeating this process for each row. If, at any row, there are no available columns that are safe, it means that the placement in the previous rows was incorrect. In that case, go back to the previous row and try placing the queen in a different column.
If you reach the last row and you are able to place a queen, then you have found a valid solution.
To find all possible solutions, continue to backtrack and try different column placements for the queens in earlier rows until all possibilities have been exhausted.

Code Implementation

def solve_n_queens(number_of_queens):
    solutions = []
    board_state = [0] * number_of_queens

    def is_safe(row_index, column_index):
        for previous_row_index in range(row_index):
            previous_column_index = board_state[previous_row_index]
            if previous_column_index == column_index:
                return False
            if abs(previous_column_index - column_index) == row_index - previous_row_index:
                return False
        return True

    def find_solutions(row_index):
        # Base case: All queens are placed.
        if row_index == number_of_queens:
            solutions.append(board_state[:])
            return

        # Try placing a queen in each column of the current row.
        for column_index in range(number_of_queens):
            if is_safe(row_index, column_index):

                # Place the queen if the position is safe
                board_state[row_index] = column_index

                # Recursively find solutions for the next row.
                find_solutions(row_index + 1)

                # Backtrack: Remove the queen and try the next column.

    find_solutions(0)
    return solutions

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
N is zero or negative	Return an empty list as there is no valid board configuration for non-positive N.
N is equal to 1	Return a list containing a single solution [['Q']] as a 1x1 board trivially allows placing one queen.
N is 2 or 3	Return an empty list, as no valid solution exists for N=2 or N=3 due to the queens being able to attack each other.
Large N values (e.g., N > 12)	The backtracking algorithm's time complexity increases exponentially, potentially leading to stack overflow or timeout errors; consider iterative solutions or pruning optimizations.
Integer overflow when calculating diagonals for very large N	Ensure data types used for diagonal calculations can handle large values, potentially using long or BigInteger if needed to prevent overflows.
Multiple valid solutions exist	The backtracking algorithm should explore all possible valid placements and collect all distinct solutions in a result list.
Memory exhaustion for large N due to storing all solutions	If memory is a constraint, consider alternative approaches like returning only the count of solutions, or generating solutions on demand without storing all of them simultaneously.
Floating point errors with approximations of diagonals.	Ensure the code relies on integer arithmetic when calculating the positions of the diagonals, rather than relying on floating point computations.