Solution Approach

The implementation uses recursive backtracking with state compression to efficiently track which cells have been filled.

State Representation:

• We use an integer array filled of length n, where filled[i] represents the state of row i
• If the j-th bit of filled[i] is 1, it means position (i, j) is filled
• This allows us to check if a cell is filled using: filled[i] >> j & 1

Main DFS Function: The dfs(i, j, t) function explores all possible tilings starting from position (i, j) with t squares already placed.

1. Base Cases:

   • If j == m: We've reached the end of a row, move to the next row by calling dfs(i + 1, 0, t)
   • If i == n: All positions are filled, update the answer with t

2. Skip Filled Positions:

   • If position (i, j) is already filled (filled[i] >> j & 1), recurse to the next position

3. Try Different Square Sizes:

   • First, find the maximum square size possible from position (i, j):
     • Count consecutive empty cells to the right (c)
     • Count consecutive empty cells downward (r)
     • Maximum square size is mx = min(r, c)

4. Place and Recurse:

   • For each possible square size w from 1 to mx:
     • Mark all cells in the w × w square as filled using bitwise OR operations:

       for k in range(w):  filled[i + w - 1] |= 1 << (j + k) # Fill bottom row of square  filled[i + k] |= 1 << (j + w - 1) # Fill right column of square

     • Recurse to the next position: dfs(i, j + w, t + 1)

5. Backtrack:

   • After exploring, restore the state by unmarking all cells in the square:

     for x in range(i, i + mx):  for y in range(j, j + mx):  filled[x] ^= 1 << y # XOR to flip bits back

Pruning: The condition t + 1 < ans ensures we only explore paths that could potentially improve our answer, pruning branches that already use too many squares.

Initialization:

• Start with ans = n * m (worst case: all 1×1 squares)
• Initialize filled = [0] * n (all cells empty)
• Call dfs(0, 0, 0) to start from the top-left corner with 0 squares placed

The algorithm systematically tries all valid square placements, using bit manipulation for efficient state tracking and pruning to avoid unnecessary exploration.

Example Walkthrough

Let's walk through tiling a 2×3 rectangle to understand the solution approach.

Initial Setup:

• Rectangle dimensions: n=2, m=3
• filled = [0, 0] (binary: [000, 000] - all cells empty)
• Start with ans = 6 (worst case: six 1×1 squares)
• Call dfs(0, 0, 0) - start at position (0,0) with 0 squares placed

Step 1: First Square Placement At position (0,0):

• Check maximum square size:
  • Consecutive empty cells right: 3
  • Consecutive empty cells down: 2
  • Maximum square size: min(2,3) = 2

Try placing a 2×2 square:

• Mark cells: filled = [6, 6] (binary: [110, 110])
• Visual state:

  [X X .] [X X .]

• Recurse to dfs(0, 2, 1) - move to position (0,2) with 1 square placed

Step 2: Second Square at (0,2) At position (0,2):

• Maximum square size possible: 1×1 (only one column left)
• Place 1×1 square at (0,2):
  • Update: filled[0] |= 1 << 2 → filled = [7, 6] (binary: [111, 110])
  • Visual state:

    [X X X] [X X .]

• Recurse to dfs(0, 3, 2) - reached end of row

Step 3: Move to Next Row Since j=3 (end of row), move to dfs(1, 0, 2)

Step 4: Process Row 1 At position (1,0):

• Cell already filled (bit check: filled[1] >> 0 & 1 = 0), skip to (1,1)

At position (1,1):

• Cell already filled, skip to (1,2)

At position (1,2):

• Empty cell found, place 1×1 square:
  • Update: filled[1] |= 1 << 2 → filled = [7, 7] (binary: [111, 111])
  • Visual state:

    [X X X] [X X X]

• Recurse to dfs(1, 3, 3)

Step 5: Complete First Solution

• Reached i=2 (all rows processed)
• Update ans = min(6, 3) = 3
• Backtrack to try other configurations

Step 6: Backtrack and Try Alternative Back at position (0,0), now try a 1×1 square instead:

• Mark only (0,0): filled = [1, 0] (binary: [001, 000])
• Visual state:

  [X . .] [. . .]

• Continue exploring...

This process would continue, trying different square sizes at each position. For example:

• 1×1 at (0,0), then 2×2 at (0,1), then 1×1 at (1,0) → 3 squares total
• Six 1×1 squares → 6 squares total

The algorithm explores all possibilities while pruning paths that exceed the current best answer. The bit manipulation (filled[i] >> j & 1) efficiently checks if cells are occupied, and the backtracking ensures we try all valid configurations to find the minimum of 3 squares.

Time and Space Complexity

Time Complexity: O((n*m)^(n*m))

The algorithm uses backtracking with pruning to explore all possible ways to tile an n×m rectangle with square tiles. In the worst case:

• Each cell in the n×m grid can be the starting point of a square tile
• For each position (i,j), we can place squares of sizes from 1×1 up to min(n-i, m-j)
• The recursion tree has a maximum depth of n*m (if we place all 1×1 tiles)
• At each level, we potentially have multiple branching factors based on the square sizes we can place
• The state space is exponential as we explore different combinations of square placements

The pruning condition t + 1 < ans helps reduce the search space, but the worst-case remains exponential.

Space Complexity: O(n + n*m)

The space complexity consists of:

• O(n) for the filled array which uses n integers to represent the filled state of the grid using bitmasks
• O(n*m) for the recursion call stack in the worst case when the maximum recursion depth equals the number of cells
• Additional O(1) space for variables like ans, r, c, mx, etc.

The dominant factor is the recursion depth, giving us O(n + n*m) which simplifies to O(n*m) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Bit Manipulation During Square Placement

The most critical pitfall in this solution is the incorrect way of marking cells as filled when placing a square. The current code only marks the bottom row and rightmost column of the square, leaving the interior cells unmarked.

Problem in Original Code:

# This only marks the edges, not the entire square! for k in range(square_size):  filled_cells[row + square_size - 1] |= 1 << (col + k) # Bottom row only  filled_cells[row + k] |= 1 << (col + square_size - 1) # Right column only

For a 3×3 square, this would only mark:

. . X . . X X X X

Instead of the entire square.

Correct Solution:

# Mark ALL cells in the square as filled for x in range(row, row + square_size):  for y in range(col, col + square_size):  filled_cells[x] |= 1 << y

2. Inconsistent Backtracking Logic

The backtracking section uses max_square_size instead of the actual square_size that was placed, leading to incorrect restoration of state.

Problem:

# Uses max_square_size for ALL squares, regardless of which size was actually placed for x in range(row, row + max_square_size):  for y in range(col, col + max_square_size):  filled_cells[x] ^= 1 << y

Correct Solution:

# Backtrack should be inside the loop, specific to each square_size for square_size in range(1, max_square_size + 1):  # Mark cells  for x in range(row, row + square_size):  for y in range(col, col + square_size):  filled_cells[x] |= 1 << y   # Recurse  backtrack(row, col + square_size, num_squares + 1)   # Backtrack for THIS specific square_size  for x in range(row, row + square_size):  for y in range(col, col + square_size):  filled_cells[x] ^= 1 << y

3. Integer Overflow with Bit Operations

For larger values of m, using bit shifting can cause issues since Python integers have practical limits for bit operations in certain contexts.

Problem: When m is large (e.g., > 30), 1 << col might behave unexpectedly in some Python implementations or be inefficient.

Solution: Add validation or use alternative state representation for very large rectangles:

def tilingRectangle(self, n: int, m: int) -> int:  # Swap dimensions if needed to minimize bit operations  if m > n:  n, m = m, n   if m > 30: # Consider alternative approach for very wide rectangles  # Use different state representation (e.g., 2D boolean array)  pass

Complete Corrected Code:

class Solution:  def tilingRectangle(self, n: int, m: int) -> int:  def backtrack(row: int, col: int, num_squares: int) -> None:  nonlocal min_squares   if col == m:  backtrack(row + 1, 0, num_squares)  return   if row == n:  min_squares = min(min_squares, num_squares)  return   if filled_cells[row] >> col & 1:  backtrack(row, col + 1, num_squares)  return   if num_squares + 1 >= min_squares:  return   # Find maximum square size  max_size = min(n - row, m - col)  for r in range(row, min(n, row + max_size)):  for c in range(col, min(m, col + max_size)):  if filled_cells[r] >> c & 1:  max_size = min(max_size, max(r - row, c - col))   # Try each square size  for size in range(1, max_size + 1):  # Mark all cells in the square  for r in range(row, row + size):  for c in range(col, col + size):  filled_cells[r] |= 1 << c   backtrack(row, col + size, num_squares + 1)   # Unmark all cells in the square  for r in range(row, row + size):  for c in range(col, col + size):  filled_cells[r] ^= 1 << c   min_squares = n * m  filled_cells = [0] * n  backtrack(0, 0, 0)  return min_squares