1240. Tiling a Rectangle with the Fewest Squares

Problem Description

The problem is asking for the smallest number of integer-sided squares that can completely cover a rectangular area without any overlap or gaps. The dimensions of the rectangle are given by n (height) and m (width). Each square used must have its sides of integer length, and the task is to find the minimum number of such squares needed to cover the entire area of the rectangle.

Intuition

To find the minimum number of squares required to tile a given rectangle, a common approach is to use recursive backtracking, which is a kind of Depth First Search (DFS) algorithm. The basic idea is to place a square in the rectangle and then recursively attempt to fill the remaining area with smaller squares. The backtracking approach ensures that all possible combinations of squares are considered.

The intuition behind this solution is to start filling the rectangle with the largest possible square from the top-left corner and recursively move forward to fill in the uncovered area. Whenever we place a square, we update our filling state and then look for the next uncovered cell to continue the process. Each time we place a square, we increase our total count by one and try to minimize this count as we search for our solution. Whenever we hit a filled cell or the end of the rectangle, we return from the recursive call, as we've either found a complete tiling or reached a point where we can't place further squares.

The algorithm can be optimized by keeping track of the filled cells using a bit mask for each row, which allows for quick updates and checks. We also keep a global variable 'ans' that holds the minimum number of squares found so far, to avoid unnecessary recursive calls whenever we exceed this number.

This approach is efficient because it incorporates a few optimizations to prune the search space:

1. If we reach the end of a row, we go to the beginning of the next row.
2. If we reach the end of the rectangle, we've found a possible tiling and update the answer if it's better than the current one.
3. We skip filled cells by moving to the next cell.
4. Before placing a new square, we calculate the maximum possible size of the square that can be placed at the current position. This is done by checking the contiguous unfilled cells to the right and below the current cell.
5. We place larger squares before smaller ones, aiming to reduce the number of squares required.
6. While backtracking, we remove the placed squares from right to left and update the filled state accordingly.

In essence, we are exploring all feasible ways to tile the rectangle with squares, aiming to do so with the fewest squares possible, leveraging recursion to represent the depth of our search and backtracking to undo steps and explore different paths.

Learn more about Backtracking patterns.

Solution Approach

The provided code implements a solution approach using recursive backtracking. Here is a step-by-step explanation of the algorithm and how the solution works:

1. A recursive depth-first search function dfs is defined within the tilingRectangle method. The dfs function accepts parameters i and j, which represent the current position (row and column) in the rectangle, and t, which is the current number of squares used to tile the rectangle so far.

2. A global variable ans is used to keep track of the minimum number of squares discovered during the recursive exploration.

3. The base cases of the dfs function are as follows:

   • If the current column j is equal to the rectangle width m, move to the start of the next row by incrementing i and resetting j to 0.
   • If the current row i is equal to the rectangle height n, which means the rectangle has been completely filled, update the global ans to the current tile count t.

4. A bitmap filled is used to represent which cells in the rectangle are already covered by squares, where each integer in the filled list corresponds to a row in the rectangle. The bits of the integer indicate whether a specific column in that row is filled.

5. The dfs function then checks if the current cell is already filled:

   • If it is filled, the function proceeds to the next cell by calling dfs(i, j + 1, t).
   • If it is not filled and the current tile count t + 1 is less than ans, the algorithm explores placing a new square in the current cell.

6. To find the largest square that can be placed at the current position, the algorithm checks contiguous unfilled cells to the right and below the current cell to determine the possible size of the square. This size is stored in mx.

7. The algorithm then places each possible square size starting from the largest (mx) down to the smallest (1). For each size w, it:

   • Marks the cells it covers as filled by updating the filled bitmap.
   • Calls dfs recursively for the next position with an updated tile count t + 1.

8. After each recursive call, backtracking occurs. The algorithm unmarks the cells of the last square placed by updating the filled bitmap to remove the square of size w, allowing it to explore different configurations.

9. Finally, after exploring all possibilities through dfs, the tilingRectangle method returns the global variable ans, which contains the minimum number of squares found to tile the rectangle.

The mentioned algorithm relies heavily on backtracking and bit manipulation to efficiently explore the search space and determine the optimal arrangement of squares. Despite the recursive nature and potentially large search space of the problem, these optimizations help manage the complexity and arrive at an answer within practical time constraints.

Example Walkthrough

Let's illustrate the solution approach using a small example where the dimensions of the rectangle are n = 2 (height) and m = 3 (width), and we are trying to find the smallest number of integer-sided squares that can cover this area.

1. Start with no squares placed, so t = 0, i = 0, and j = 0. The filled bitmap is initially all zeros, indicating no cells are covered.

2. Since we're at the top-left corner (cell 0,0), we want to place the largest square possible. Here, the largest square we can place is 2x2, because placing a 3x3 would exceed the rectangle's width.

3. Place a 2x2 square which updates t to 1. The filled bitmap is updated to show that the first two rows and first two columns are covered.

   filled after placing 2x2 square:

   • Row 0: 110 (in binary)
   • Row 1: 110 (in binary)

4. With the current cell (0,0) filled, we move to the first unfilled cell, which is (0,2).

5. At position (0,2), the largest square we can place is a 1x1 because of the rectangle's edge. We update t to 2 and place the square.

   filled after placing 1x1 square:

   • Row 0: 111 (in binary)
   • Row 1: 110 (in binary)

6. All cells in row 0 are now filled, so we move to the next row, position (1,2).

7. Again, at position (1,2), the largest square we can place is 1x1. We update t to 3 and place the square.

   filled after placing another 1x1 square:

   • Row 0: 111 (in binary)
   • Row 1: 111 (in binary)

8. The entire rectangle is now filled, and t = 3. Since this is the first solution found, ans is updated to 3.

9. No further recursive calls are necessary as we have filled the entire rectangle, and we have updated the ans to the minimum number of squares.

10. The algorithm ends, and we return ans, which is 3 for this example.

Thus, for a rectangle of size 2x3, the smallest number of integer-sided squares required to cover it without overlap or gaps is 3, consisting of one 2x2 square and two 1x1 squares.

Solution Implementation

Time and Space Complexity

The provided code is a backtracking algorithm to solve the problem of tiling a n x m rectangle with squares of different sizes. The algorithm performs a depth-first search (DFS) to find the smallest number of tiles required.

Time Complexity

The time complexity of the algorithm is hard to determine precisely due to the nature of the problem, but we can break down the complexity into different parts:

1. DFS Calls: The DFS function is called recursively, attempting to fit squares of all possible sizes starting from the current largest possible (determined by mx variable). For each call to dfs, it is possible to place a square at the current position (i, j) and the range of squares is from 1x1 to mx x mx, where mx is the min possible size decided by the current state of filled array, i.e., smaller of the two sides from the current position to the edges of the unfilled region.

2. Inner Loops: Within each call, there are two loops to determine how big a square can be placed at position (i, j). The time complexity for these loops, in the worst case, would be O(n) for the row check and O(m) for the column check, where n and m are the dimensions of the rectangle.

3. Square Placement & Removal: Squares are placed and removed from the filled array using bitwise operations. These involve O(mx^2) operations for placement and O(w) operations for removal per depth.

Considering all the factors, the time complexity is exponential as it depends on the rectangle dimensions and the placement possibilities. The upper bound of this can be approximated as O(n * m * min(n, m)^(n * m)), but this is a very loose upper bound because the algorithm has several early stopping conditions (e.g., the check if t + 1 < ans) which can greatly reduce the search space.

Space Complexity

The space complexity is easier to ascertain:

1. Recursion Stack: Since the algorithm uses DFS, the recursion stack height could go as deep as the total number of squares put into the rectangle. This is in the worst-case O(n * m).

2. Filled Array: The state of the rectangle is maintained using an array filled that is O(n).

Thus, the space complexity is O(n * m) due to the recursion stack. The additional O(n) used by filled does not dominate the space complexity.

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