Problem Description

You are given a rectangle with dimensions n x m. Your task is to completely cover this rectangle using square tiles, where each square tile must have integer side lengths. The goal is to find the minimum number of square tiles needed to tile the entire rectangle.

For example, if you have a 2x3 rectangle, you could tile it with 6 squares of size 1x1, but a better solution would be to use 2 squares of size 2x2 and 2 squares of size 1x1, for a total of 3 squares.

The squares can be of different sizes, and you need to arrange them in such a way that:

• The entire rectangle is covered
• There are no overlapping squares
• There are no gaps between squares
• The total number of squares used is minimized

The function should return the minimum number of integer-sided squares required to tile the given n x m rectangle.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem involves tiling a rectangle with squares, not nodes and edges as in graph problems.

Need to solve for kth smallest/largest?

• No: We're looking for the minimum number of squares needed, not the kth smallest or largest element.

Involves Linked Lists?

• No: The problem deals with a 2D rectangle grid, not linked lists.

Does the problem have small constraints?

• Yes: The problem typically has small constraints (n, m ≤ 13) because we need to explore different ways to tile the rectangle. The search space grows exponentially, making it impractical for large inputs.

Brute force / Backtracking?

• Yes: We need to try different square placements and backtrack when a configuration doesn't lead to an optimal solution. The solution involves:
  • Recursively trying to place squares of different sizes at each empty position
  • Tracking which positions have been filled using state compression
  • Backtracking by removing placed squares to try alternative configurations
  • Pruning branches that exceed the current minimum number of squares

Conclusion: The flowchart correctly identifies that this is a backtracking problem due to the small constraints and the need to explore all possible ways to tile the rectangle optimally.

Intuition

When faced with tiling a rectangle with the minimum number of squares, we need to think about how to systematically explore all possible ways to place squares.

The key insight is that we can process the rectangle position by position, from left to right and top to bottom. At each empty position, we have choices - we can place squares of different sizes starting from that position. For instance, if we're at position (i, j) and have a 3x3 empty space available, we could place a 1x1, 2x2, or 3x3 square there.

Why backtracking? Because we don't know which choice will lead to the optimal solution until we try them all. After placing a square, we move to the next empty position and repeat the process. If we reach a configuration that uses more squares than our current best solution, we can stop exploring that branch (pruning). When we finish one path, we backtrack by removing the last placed square and trying a different size.

The state compression technique comes from realizing that we only need to know which cells are filled or empty. Since each cell has only two states (filled/empty), we can use bits to represent them. For each row, we use an integer where the j-th bit indicates whether position (i, j) is filled. This makes checking and updating the state very efficient using bitwise operations.

The recursive approach naturally handles the exploration:

• Start from position (0, 0)
• For each empty position, try all possible square sizes that fit
• Mark the covered cells, recurse to handle the rest
• When backtracking, unmark those cells to try other options
• Keep track of the minimum number of squares needed

This exhaustive search with pruning ensures we find the optimal solution despite the exponential search space, which is manageable due to the small constraints.

Learn more about Backtracking patterns.

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.

Solution Implementation

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