DSA Notes
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Master the N-Queens problem completely — problem definition, why queens attack each other, step-by-step backtracking solution, constraint checking using column and diagonal sets, optimized solution with bitmasks, counting total solutions, and visualizing the complete search tree.
Problem Statement
Place N queens on an N×N chessboard such that no two queens attack each other. A queen attacks all squares in the same row, same column, and both diagonals.
Example
Valid 4-Queens solution:
. Q . .
. . . Q
Q . . .
. . Q .
(Q = queen, . = empty)
No two queens share a row, column, or diagonal.
Solution 1 — Basic Backtracking
Example
def solve_n_queens(n):
"""
Place n queens on n×n board with no two attacking each other.
Returns list of all valid board configurations.
Each configuration is a list of column positions for each row.
"""
results = []
queens = [] # queens[row] = column where queen is placed in that row
def is_safe(row, col):
"""Check if placing a queen at (row, col) is safe."""
for r in range(row):
c = queens[r]
if c == col: # Same column
return False
if abs(r - row) == abs(c - col): # Same diagonal
return False
return True
def backtrack(row):
if row == n:
# Convert queens array to board representation
board = []
for r in range(n):
row_str = '.' * queens[r] + 'Q' + '.' * (n - queens[r] - 1)
board.append(row_str)
results.append(board)
return
for col in range(n):
if is_safe(row, col):
queens.append(col) # CHOOSE: place queen at (row, col)
backtrack(row + 1) # EXPLORE: place queen in next row
queens.pop() # UNDO: remove queen from (row, col)
backtrack(0)
return results
solutions = solve_n_queens(4)
print(f"Number of solutions for 4-Queens: {len(solutions)}") # 2
for sol in solutions:
for row in sol:
print(row)
print()
Output for 4-Queens:
Example
.Q.. ..Q.
...Q Q...
Q... ...Q
..Q. .Q..
Solution 2 — Optimized with Hash Sets: O(n!)
The is_safe function in Solution 1 takes O(n) time. Using sets for column and diagonal tracking makes it O(1):
Example
def solve_n_queens_fast(n):
"""
Optimized N-Queens using sets for O(1) conflict checking.
Time: O(n!), Space: O(n)
"""
results = []
queens = [-1] * n
# Track which columns and diagonals are occupied
cols = set()
diag1 = set() # For diagonal: r - c is constant along ↘ diagonal
diag2 = set() # For anti-diagonal: r + c is constant along ↗ diagonal
def backtrack(row):
if row == n:
board = []
for r in range(n):
board.append('.' * queens[r] + 'Q' + '.' * (n - queens[r] - 1))
results.append(board)
return
for col in range(n):
# O(1) conflict check using sets
if col in cols or (row - col) in diag1 or (row + col) in diag2:
continue # PRUNE: This position is under attack
# CHOOSE
queens[row] = col
cols.add(col)
diag1.add(row - col)
diag2.add(row + col)
# EXPLORE
backtrack(row + 1)
# UNDO
cols.remove(col)
diag1.remove(row - col)
diag2.remove(row + col)
backtrack(0)
return results
print(f"4-Queens: {len(solve_n_queens_fast(4))} solutions") # 2
print(f"8-Queens: {len(solve_n_queens_fast(8))} solutions") # 92
Trace — 4-Queens Solution
| Board | 4×4, placing queens row by row |
| Row 0 | Try col 0 |
| Row 1: col 0 | same col, skip |
| col 1 | diag1: 1-1=0 already in diag1, skip |
| col 2 | 1-2=-1, 1+2=3 — SAFE! Place |
| Row 2: col 0 | col in cols, skip |
| col 1 | 2+1=3 in diag2, skip |
| col 2 | col in cols, skip |
| col 3 | 2-3=-1 in diag1, skip |
| No valid placement | BACKTRACK |
| Row 1: col 3 | 1-3=-2, 1+3=4 — SAFE! Place |
| Row 2: col 0 | col in cols, skip |
| col 1 | 2-1=1, 2+1=3 — SAFE! Place |
| Row 3: col 0 | col in cols, skip |
| col 1 | col in cols, skip |
| col 2 | 3-2=1 in diag1, skip |
| col 3 | col in cols, skip |
| Row 2: col 2 | 2-2=0 in diag1, skip |
| col 3 | col in cols, skip |
| Row 0: Try col 1 | leads to first solution: [1,3,0,2] |
Solution 3 — Count Only (Bitmask): O(n!) — Fastest
For large n, if you only need the count (not the boards), bitmasks are fastest:
python example
def count_n_queens(n):
"""
Count N-Queens solutions using bitmasks.
Dramatically faster for large n because bitwise ops are single instructions.
Time: O(n!), but with much smaller constant than set-based approach.
"""
all_ones = (1 << n) - 1 # n bits set to 1 (all columns)
count = 0
def backtrack(cols, diag1, diag2):
nonlocal count
if cols == all_ones: # All columns filled → found a solution
count += 1
return
# Available positions: columns not blocked by any queen
available = all_ones & ~(cols | diag1 | diag2)
while available:
# Pick the rightmost available position
position = available & (-available) # Isolate lowest set bit
available &= available - 1 # Clear lowest set bit
backtrack(
cols | position,
(diag1 | position) << 1, # Diagonal shifts left as row increases
(diag2 | position) >> 1 # Anti-diagonal shifts right
)
backtrack(0, 0, 0)
return count
for n in range(1, 10):
print(f"{n}-Queens: {count_n_queens(n)} solutions")
# 1-Queens: 1
# 2-Queens: 0
# 3-Queens: 0
# 4-Queens: 2
# 5-Queens: 10
# 6-Queens: 4
# 7-Queens: 40
# 8-Queens: 92
# 9-Queens: 352N-Queens Solutions Table
| N | Solutions |
|---|---|
| 1 | 1 |
| 2 | 0 |
| 3 | 0 |
| 4 | 2 |
| 5 | 10 |
| 6 | 4 |
| 7 | 40 |
| 8 | 92 |
| 9 | 352 |
| 10 | 724 |
| 11 | 2,680 |
| 12 | 14,200 |
| 13 | 73,712 |
First Non-Attacking Queen Placement
Example
def first_n_queens(n):
"""Return just the first valid solution found."""
queens = [-1] * n
cols, diag1, diag2 = set(), set(), set()
def backtrack(row):
if row == n:
return True # Found a solution
for col in range(n):
if col in cols or (row-col) in diag1 or (row+col) in diag2:
continue
queens[row] = col
cols.add(col); diag1.add(row-col); diag2.add(row+col)
if backtrack(row + 1):
return True # Propagate success, don't backtrack
queens[row] = -1
cols.remove(col); diag1.remove(row-col); diag2.remove(row+col)
return False # No valid placement found for this row
if backtrack(0):
return queens
return None
print(first_n_queens(8)) # [0, 4, 7, 5, 2, 6, 1, 3] — one valid solution
Complexity Analysis
| Approach | Time | Space | Notes |
|---|---|---|---|
| Basic (O(n) conflict check) | O(n! × n) | O(n) | Slowest |
| Set-based (O(1) conflict check) | O(n!) | O(n) | Standard interview |
| Bitmask | O(n!) | O(n) | Fastest in practice |
All approaches are ultimately O(n!) because there are O(n!) possible permutations of queen column positions to check. The optimization is in the constant factor.
Summary
The N-Queens problem is the canonical backtracking problem because it has:
- One decision per step: Which column to place the queen in (for the current row)
- Clear validity check: No two queens can share column or diagonal
- Perfect undo semantics: Remove queen, remove from sets
- Effective pruning: O(1) conflict detection eliminates most branches early
The three approaches (basic, set-based, bitmask) demonstrate the progression from correct solution to optimized solution — a pattern applicable to all backtracking problems.
*Next Lesson: Rat in a Maze*
Exam Focus
Revise definitions, diagrams, examples, and short-answer points for N-Queens Problem — Complete Solution with Backtracking and Optimization.
Interview Use
Prepare one clear explanation, one practical example, and one common mistake for this Data Structures & Algorithms topic.
Search Terms
data-structures-algorithms, data structures & algorithms, data, structures, algorithms, backtracking, queens, problem
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