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.
Flowchart Walkthrough
Let's use the algorithm flowchart to analyze and deduce the problem with the Backtracking pattern for LeetCode 1240, "Tiling a Rectangle with the Fewest Squares". You can view the complete decision-making process on the Flowchart. Here’s how we proceed:
-
Is it a graph?
- No: The problem does not involve a graph with nodes and edges in the conventional sense but deals with tiling a 2D area.
-
Need to solve for kth smallest/largest?
- No: The problem asks for the minimum number of squares to tile a rectangle, not the kth smallest or largest.
-
Involves Linked Lists?
- No: This problem does not involve data structures like linked lists.
-
Does the problem have small constraints?
- Yes: The problem states restrictions on rectangle dimensions (not exceeding 13x13) which suggests a smaller state space manageable for more exhaustive techniques like backtracking.
-
Brute force / Backtracking?
- Yes: Given the small constraints, a brute-force approach or backtracking is feasible. The problem involves recursively trying to place different square sizes within the rectangle and checking for the optimal configuration, fitting the definition of backtracking where multiple decisions are explored to seek an optimal solution.
Conclusion: Following the Flowchart, backtracking is recommended due to the problem’s nature of finding an optimal configuration (minimum number of squares) within a bounded area, supported by the small constraints allowing the exploration of possible configurations exhaustively.
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:
- If we reach the end of a row, we go to the beginning of the next row.
- 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.
- We skip filled cells by moving to the next cell.
- 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.
- We place larger squares before smaller ones, aiming to reduce the number of squares required.
- 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:
-
A recursive depth-first search function
dfs
is defined within thetilingRectangle
method. Thedfs
function accepts parametersi
andj
, which represent the current position (row and column) in the rectangle, andt
, which is the current number of squares used to tile the rectangle so far. -
A global variable
ans
is used to keep track of the minimum number of squares discovered during the recursive exploration. -
The base cases of the
dfs
function are as follows:- If the current column
j
is equal to the rectangle widthm
, move to the start of the next row by incrementingi
and resettingj
to 0. - If the current row
i
is equal to the rectangle heightn
, which means the rectangle has been completely filled, update the globalans
to the current tile countt
.
- If the current column
-
A bitmap
filled
is used to represent which cells in the rectangle are already covered by squares, where each integer in thefilled
list corresponds to a row in the rectangle. The bits of the integer indicate whether a specific column in that row is filled. -
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 thanans
, the algorithm explores placing a new square in the current cell.
- If it is filled, the function proceeds to the next cell by calling
-
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
. -
The algorithm then places each possible square size starting from the largest (
mx
) down to the smallest (1). For each sizew
, it:- Marks the cells it covers as filled by updating the
filled
bitmap. - Calls
dfs
recursively for the next position with an updated tile countt + 1
.
- Marks the cells it covers as filled by updating the
-
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 sizew
, allowing it to explore different configurations. -
Finally, after exploring all possibilities through
dfs
, thetilingRectangle
method returns the global variableans
, 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.
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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.
-
Start with no squares placed, so
t = 0
,i = 0
, andj = 0
. Thefilled
bitmap is initially all zeros, indicating no cells are covered. -
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 a3x3
would exceed the rectangle's width. -
Place a
2x2
square which updatest
to 1. Thefilled
bitmap is updated to show that the first two rows and first two columns are covered.filled
after placing2x2
square:- Row 0: 110 (in binary)
- Row 1: 110 (in binary)
-
With the current cell (0,0) filled, we move to the first unfilled cell, which is (0,2).
-
At position (0,2), the largest square we can place is a
1x1
because of the rectangle's edge. We updatet
to 2 and place the square.filled
after placing1x1
square:- Row 0: 111 (in binary)
- Row 1: 110 (in binary)
-
All cells in row 0 are now filled, so we move to the next row, position (1,2).
-
Again, at position (1,2), the largest square we can place is
1x1
. We updatet
to 3 and place the square.filled
after placing another1x1
square:- Row 0: 111 (in binary)
- Row 1: 111 (in binary)
-
The entire rectangle is now filled, and
t = 3
. Since this is the first solution found,ans
is updated to 3. -
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. -
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
1class Solution:
2 def tilingRectangle(self, n: int, m: int) -> int:
3
4 # Define depth-first search function
5 def dfs(row: int, col: int, tiles_used: int):
6 nonlocal min_tiles # Access the 'min_tiles' in the enclosing 'tilingRectangle' method
7 # Move to the next row if the end of the current row is reached
8 if col == m:
9 row += 1
10 col = 0
11 # Update the minimum number of tiles if the end of the rectangle is reached
12 if row == n:
13 min_tiles = tiles_used
14 return
15 # Continue with the next column if the current position is already filled
16 if filled[row] >> col & 1:
17 dfs(row, col + 1, tiles_used)
18 elif tiles_used + 1 < min_tiles: # Only continue if it is possible to use fewer tiles than the current minimum
19 max_row, max_col = 0, 0
20 # Find how far we can extend the tile downwards
21 for k in range(row, n):
22 if filled[k] >> col & 1:
23 break
24 max_row += 1
25 # Find how far we can extend the tile to the right
26 for k in range(col, m):
27 if filled[row] >> k & 1:
28 break
29 max_col += 1
30 # The maximum size of the square tile we can place
31 max_tile_size = min(max_row, max_col)
32
33 for x in range(row, row + max_tile_size):
34 for y in range(col, col + max_tile_size):
35 filled[x] |= 1 << y # Mark the square as filled
36 # Attempt to place every possible square tile
37 for width in range(max_tile_size, 0, -1):
38 dfs(row, col + width, tiles_used + 1) # Recurse with one more tile
39 # Undo the tile placement
40 for k in range(width):
41 filled[row + width - 1] ^= 1 << (col + k)
42 if k < width - 1:
43 filled[row + k] ^= 1 << (col + width - 1)
44
45 min_tiles = n * m # Initialize the minimum number of tiles required with the maximum possible number
46 filled = [0] * n # Represent each row with a bitmap to show if a cell is filled
47 dfs(0, 0, 0) # Start the depth-first search from the top-left corner
48 return min_tiles # Return the minimum number of tiles found
49
1class Solution {
2 private int numRows; // Number of rows in the rectangle
3 private int numCols; // Number of columns in the rectangle
4 private int[] tileCovered; // Bit representation of tiles covered in each row
5 private int minTiles; // Minimum number of tiles needed to cover the rectangle
6
7 public int tilingRectangle(int n, int m) {
8 numRows = n;
9 numCols = m;
10 minTiles = n * m; // Initial maximum is the total area, which is the worst case
11 tileCovered = new int[n];
12 fillRectangle(0, 0, 0);
13 return minTiles;
14 }
15
16 private void fillRectangle(int row, int col, int tilesUsed) {
17 if (col == numCols) {
18 row++; // Move to the next row
19 col = 0; // Reset column to the first one
20 }
21 if (row == numRows) {
22 minTiles = tilesUsed; // Update the minimum tiles used
23 return;
24 }
25 if ((tileCovered[row] >> col & 1) == 1) {
26 fillRectangle(row, col + 1, tilesUsed); // If the tile at (row, col) is covered, move to next tile
27 } else if (tilesUsed + 1 < minTiles) {
28 int maxRow = 0, maxCol = 0;
29 // Calculate how far we can extend squares to the right and downwards
30 for (int i = row; i < numRows && (tileCovered[i] >> col & 1) == 0; ++i) {
31 maxRow++;
32 }
33 for (int j = col; j < numCols && (tileCovered[row] >> j & 1) == 0; ++j) {
34 maxCol++;
35 }
36 int maxSize = Math.min(maxRow, maxCol); // Maximum size of the square we can place
37
38 // Mark the tiles as covered for the largest square initially
39 for (int x = row; x < row + maxSize; ++x) {
40 for (int y = col; y < col + maxSize; ++y) {
41 tileCovered[x] |= 1 << y;
42 }
43 }
44
45 // Try placing squares of all possible sizes starting from largest to smallest
46 for (int squareSize = maxSize; squareSize > 0; --squareSize) {
47 fillRectangle(row, col + squareSize, tilesUsed + 1);
48
49 // Uncover tiles for a square of the current size
50 for (int k = 0; k < squareSize; ++k) {
51 tileCovered[row + squareSize - 1] ^= 1 << (col + k);
52 if (k < squareSize - 1) {
53 tileCovered[row + k] ^= 1 << (col + squareSize - 1);
54 }
55 }
56 }
57 }
58 }
59}
60
1#include<cstring>
2
3class Solution {
4public:
5 int tilingRectangle(int n, int m) {
6 memset(isFilled, 0, sizeof(isFilled)); // Initialize all tiles as unfilled
7 this->height = n;
8 this->width = m;
9 minimumTiles = height * width; // Start with the worst-case: using all 1x1 tiles
10 dfs(0, 0, 0); // Begin the depth-first search at the top-left corner of the rectangle
11 return minimumTiles; // Return the minimal number of tiles that can fill the rectangle
12 }
13
14private:
15 int isFilled[13]; // Represents the filled status of each row up to 13, which is the maximum size
16 int height, width;
17 int minimumTiles;
18
19 void dfs(int row, int col, int tileCount) {
20 if (col == width) { // We've reached the end of a row
21 row++; // Move to the next row
22 col = 0; // Start from the first column of the new row
23 }
24 if (row == height) { // If every row is covered
25 minimumTiles = tileCount; // Update the minimum number of tiles
26 return;
27 }
28 if ((isFilled[row] >> col) & 1) { // If the current tile is already filled
29 dfs(row, col + 1, tileCount); // Move to the next column
30 } else if (tileCount + 1 < minimumTiles) { // If we could still potentially beat the current best
31 int maxRowExtent = 0, maxColExtent = 0;
32 for (int r = row; r < height; ++r) { // Calculate maximum extend in the rows
33 if ((isFilled[r] >> col) & 1)
34 break;
35 maxRowExtent++;
36 }
37 for (int c = col; c < width; ++c) { // Calculate maximum extend in the columns
38 if ((isFilled[row] >> c) & 1)
39 break;
40 maxColExtent++;
41 }
42
43 // Limit by the smaller dimension to identify the maximum size of square tile we can use
44 int maxSize = min(maxRowExtent, maxColExtent);
45
46 // Mark the filled tiles within the square defined by maxSize
47 for (int x = row; x < row + maxSize; ++x) {
48 for (int y = col; y < col + maxSize; ++y) {
49 isFilled[x] |= 1 << y;
50 }
51 }
52
53 // Try to place a square tile and continue the search
54 for (int squareSize = maxSize; squareSize > 0; --squareSize) {
55 dfs(row, col + squareSize, tileCount + 1);
56
57 // Unfill the tiles on the last row and column of the square to backtrack
58 for (int k = 0; k < squareSize; ++k) {
59 isFilled[row + squareSize - 1] ^= 1 << (col + k);
60 if (k < squareSize - 1) {
61 isFilled[row + k] ^= 1 << (col + squareSize - 1);
62 }
63 }
64 }
65 }
66 }
67};
68
1function tilingRectangle(n: number, m: number): number {
2 let minimumTiles = n * m; // Initialize the minimum number of tiles needed
3 const isFilled: number[] = new Array(n).fill(0); // Track filled cells in the rectangle
4
5 // Depth-first search to find the minimum number of tiles needed
6 const search = (row: number, col: number, tilesUsed: number) => {
7 // Move to the next row if the current one is filled
8 if (col === m) {
9 row++;
10 col = 0;
11 }
12 // Update the answer if we've filled the entire rectangle
13 if (row === n) {
14 minimumTiles = tilesUsed;
15 return;
16 }
17 // Skip already filled cells
18 if ((isFilled[row] >> col) & 1) {
19 search(row, col + 1, tilesUsed);
20 return;
21 }
22 // Only continue if we can potentially use fewer tiles than current minimumTiles
23 if (tilesUsed + 1 < minimumTiles) {
24 let maxRow = 0, maxCol = 0;
25 // Find the maximum size of the square we can place
26 for (let r = row; r < n; ++r) {
27 if ((isFilled[r] >> col) & 1) break;
28 maxRow++;
29 }
30 for (let c = col; c < m; ++c) {
31 if ((isFilled[row] >> c) & 1) break;
32 maxCol++;
33 }
34 const maxSize = Math.min(maxRow, maxCol); // Maximum tile size is limited by both dimensions
35
36 // Mark the cells as filled for the largest possible square
37 for (let x = row; x < row + maxSize; ++x) {
38 for (let y = col; y < col + maxSize; ++y) {
39 isFilled[x] |= 1 << y;
40 }
41 }
42
43 // Try to find an optimum by trying all possible square sizes
44 for (let size = maxSize; size > 0; --size) {
45 search(row, col + size, tilesUsed + 1); // Recursively search with this square placed
46 // Unfill the cells used by the current square and decrement size
47 for (let k = 0; k < size; ++k) {
48 isFilled[row + size - 1] ^= 1 << (col + k);
49 if (k < size - 1) {
50 isFilled[row + k] ^= 1 << (col + size - 1);
51 }
52 }
53 }
54 }
55 };
56
57 search(0, 0, 0); // Start searching from the top-left cell with 0 tiles used
58 return minimumTiles; // Return the minimum number of tiles needed to fill the rectangle
59}
60
61// Example:
62// console.log(tilingRectangle(2, 3)); // Output would be the minimum number of tiles needed.
63
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:
-
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 todfs
, it is possible to place a square at the current position(i, j)
and the range of squares is from1x1
tomx x mx
, wheremx
is the min possible size decided by the current state offilled
array, i.e., smaller of the two sides from the current position to the edges of the unfilled region. -
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 beO(n)
for the row check andO(m)
for the column check, wheren
andm
are the dimensions of the rectangle. -
Square Placement & Removal: Squares are placed and removed from the
filled
array using bitwise operations. These involveO(mx^2)
operations for placement andO(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:
-
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)
. -
Filled Array: The state of the rectangle is maintained using an array
filled
that isO(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.
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