N Queens Problem in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of N Queens Solution Count

What is the output of the following code that counts the number of solutions for the 4-Queens problem?

DSA Python

def is_safe(board, row, col, n):
    for i in range(col):
        if board[row][i] == 1:
            return False
    i, j = row, col
    while i >= 0 and j >= 0:
        if board[i][j] == 1:
            return False
        i -= 1
        j -= 1
    i, j = row, col
    while i < n and j >= 0:
        if board[i][j] == 1:
            return False
        i += 1
        j -= 1
    return True

def solve_n_queens(board, col, n):
    if col == n:
        return 1
    count = 0
    for i in range(n):
        if is_safe(board, i, col, n):
            board[i][col] = 1
            count += solve_n_queens(board, col + 1, n)
            board[i][col] = 0
    return count

n = 4
board = [[0]*n for _ in range(n)]
print(solve_n_queens(board, 0, n))

A 4

B 1

C 0

D 2

💡 Hint

Recall the known number of solutions for 4 queens on a 4x4 board.

The 4-Queens problem has exactly 2 distinct solutions. The code counts all valid placements recursively.

🧠 Conceptual

intermediate

Understanding Diagonal Attacks in N Queens

In the N Queens problem, why do we check diagonals when placing a queen?

A Because diagonals are irrelevant to queen placement safety.

B Because queens can attack any piece in the same diagonal direction.

C Because queens cannot attack diagonally, so we skip checking them.

D Because diagonals are the only places queens can move horizontally.

💡 Hint

Think about how a queen moves in chess.

Queens attack horizontally, vertically, and diagonally. Checking diagonals ensures no queen can attack another diagonally.

🔧 Debug

advanced

Identify the Error in N Queens Safety Check

What error will this code produce when checking if a queen can be placed safely?

DSA Python

def is_safe(board, row, col, n):
    for i in range(col):
        if board[row][i] == 1:
            return False
    i, j = row, col
    while i >= 0 and j >= 0:
        if board[i][j] == 1:
            return False
        i -= 1
        j -= 1
    i, j = row, col
    while i < n and j >= 0:
        if board[i][j] == 1:
            return False
        i += 1
        j -= 1
    return True

board = [[0]*4 for _ in range(4)]
print(is_safe(board, 4, 2, 4))

A Returns True

B Returns False

C IndexError: list index out of range

D SyntaxError

💡 Hint

Check the row index passed to is_safe.

Row index 4 is outside the 0-3 range for a 4x4 board, causing IndexError.

🚀 Application

advanced

Number of Solutions for 5 Queens

How many distinct solutions does the 5-Queens problem have?

A 10

B 2

C 4

D 8

💡 Hint

Try to recall or research the known solution count for 5 queens.

The 5-Queens problem has exactly 10 distinct solutions.

❓ Predict Output

expert

Output of N Queens Board Representation

What is the printed board after placing queens for n=4 in this code?

DSA Python

def print_board(board):
    for row in board:
        print(' '.join('Q' if x == 1 else '.' for x in row))

board = [
    [0, 0, 1, 0],
    [1, 0, 0, 0],
    [0, 0, 0, 1],
    [0, 1, 0, 0]
]
print_board(board)

A '. . Q .\nQ . . .\n. . . Q\n. Q . .'

B '. Q . .\n. . . Q\nQ . . .\n. . Q .'

C 'Q . . .\n. Q . .\n. . Q .\n. . . Q'

D '. . . Q\n. Q . .\n. . Q .\nQ . . .'

💡 Hint

Look carefully at the positions of 1s in the board list.

The code prints 'Q' where the board has 1 and '.' where it has 0, row by row.

Practice

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

easy

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

B. Count the total number of queens on a chessboard

C. Move queens to capture all opponent pieces

D. Arrange queens in a single row on the chessboard

Solution

1. Step 1: Understand the problem statement

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

2. Step 2: Analyze the options

   Only Place N queens on an NxN chessboard so that no two queens attack each other correctly describes placing queens so none attack each other. Other options describe unrelated tasks.

3. Final Answer:

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

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 be in the same row or column
• Confusing the problem with counting queens
• Assuming queens can attack each other

2. Which of the following Python code snippets correctly initializes an empty chessboard for the N Queens Problem with size n?

easy

A. board = [[0 for _ in range(n)] for _ in range(n)]

B. board = [[0]*n]

C. board = [0]*n

D. board = [[0]*n]*n

Solution

1. Step 1: Understand board initialization

   We need a 2D list with independent rows to represent the chessboard.

2. Step 2: Analyze each option

   board = [[0]*n]*n creates rows referencing the same list, causing bugs. board = [[0 for _ in range(n)] for _ in range(n)] creates independent rows correctly. board = [0]*n creates a 1D list, not 2D. board = [[0]*n] creates a single row only.

3. Final Answer:

   board = [[0 for _ in range(n)] for _ in range(n)] -> Option A

4. Quick Check:

   Independent rows = board = [[0 for _ in range(n)] for _ in range(n)] [OK]

Hint: Use nested list comprehension for independent rows [OK]

Common Mistakes:

• Using [[0]*n]*n causing shared references
• Creating 1D list instead of 2D
• Forgetting to repeat for each row

3. What will be the output of the following Python code snippet for n = 4?

def is_safe(board, row, col):
    for i in range(col):
        if board[row][i] == 1:
            return False
    for i, j in zip(range(row, -1, -1), range(col, -1, -1)):
        if board[i][j] == 1:
            return False
    for i, j in zip(range(row, len(board), 1), range(col, -1, -1)):
        if board[i][j] == 1:
            return False
    return True

board = [[0]*4 for _ in range(4)]
board[1][0] = 1
print(is_safe(board, 1, 1))
print(is_safe(board, 0, 1))

medium

A. True\nFalse

B. False\nFalse

C. False\nTrue

D. True\nTrue

Solution

1. Step 1: Check is_safe for position (1,1)

   Column check: board[1][0] == 1, so returns False.

2. Step 2: Check is_safe for position (0,1)

   Row left: board[0][0] == 0. Upper diagonal: board[0][1] == 0. Lower diagonal: board[1][0] == 1, so returns False.

3. Final Answer:

   False\nFalse -> Option B

4. Quick Check:

   Check columns and diagonals = False, False [OK]

Hint: Check columns and diagonals left of position [OK]

Common Mistakes:

• Ignoring diagonal checks
• Confusing row and column indices
• Assuming safe if no queen in same row only

4. Identify the error in the following code snippet for solving the N Queens Problem:

def solve_n_queens(board, col):
    if col >= len(board):
        return True
    for row in range(len(board)):
        if is_safe(board, row, col):
            board[row][col] = 1
            if solve_n_queens(board, col + 1):
                return True
            board[row][col] = 0
    return False

board = [[0]*4 for _ in range(4)]
solve_n_queens(board, 0)
print(board)

medium

A. The base case condition is incorrect

B. The board is not reset properly after backtracking

C. The function returns True too early, missing all solutions

D. The is_safe function is not called

Solution

1. Step 1: Analyze the recursion and return statements

   The function returns True immediately after finding the first solution, so it stops searching further.

2. Step 2: Understand impact on solution finding

   This means it finds only one solution and does not explore all possible solutions.

3. Final Answer:

   The function returns True too early, missing all solutions -> Option C

4. Quick Check:

   Early return stops full search [OK]

Hint: Return True only after exploring all columns [OK]

Common Mistakes:

• Stopping recursion after first solution
• Not resetting board on backtrack
• Incorrect base case check

5. Given the N Queens Problem, which approach best improves efficiency by reducing unnecessary checks during queen placement?

hard

A. Place queens randomly and retry if conflicts occur

B. Check the entire board for conflicts every time before placing a queen

C. Use nested loops to try all possible board configurations without pruning

D. Use backtracking with arrays to track columns and diagonals already occupied

Solution

1. Step 1: Understand efficiency in N Queens

   Checking the entire board repeatedly wastes time; pruning reduces checks.

2. Step 2: Analyze options for pruning

   Using arrays to track columns and diagonals occupied avoids repeated scanning, improving speed.

3. Final Answer:

   Use backtracking with arrays to track columns and diagonals already occupied -> Option D

4. Quick Check:

   Pruning with tracking arrays = Use backtracking with arrays to track columns and diagonals already occupied [OK]

Hint: Track columns and diagonals to skip invalid spots fast [OK]

Common Mistakes:

• Checking full board every time
• Trying random placements without pruning
• Using brute force without backtracking