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Tiling a Rectangle with the Fewest Squares

Tiling a Rectangle with the Fewest Squares problem asks for the minimum number of squares required to cover a rectangle using integer-sided squares.

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Hard · Backtracking search with pruning

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Answer-first summary

Tiling a Rectangle with the Fewest Squares problem asks for the minimum number of squares required to cover a rectangle using integer-sided squares.

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This problem requires finding the minimum number of integer-sided squares to tile a rectangle of size n x m. A backtracking approach with pruning is often used to explore possible configurations efficiently, pruning redundant paths. The goal is to minimize the number of squares required while filling the entire area of the rectangle.

Problem Statement

You are given a rectangle of size n x m. Your task is to find the minimum number of integer-sided squares that can tile the rectangle without overlapping. The rectangles can be tiled with squares of varying sizes, and the solution must cover the entire area without any gaps.

For example, with n = 2 and m = 3, three squares are needed: two 1x1 squares and one 2x2 square. The challenge lies in identifying the fewest squares needed and applying an efficient backtracking approach to explore possible configurations.

Examples

Example 1

Input: n = 2, m = 3

Output: 3

3 squares are necessary to cover the rectangle. 2 (squares of 1x1) 1 (square of 2x2)

Example 2

Input: n = 5, m = 8

Output: 5

Example details omitted.

Example 3

Input: n = 11, m = 13

Output: 6

Example details omitted.

Constraints

  • 1 <= n, m <= 13

Solution Approach

Backtracking Search

A backtracking approach is used to recursively place squares on the rectangle. The search explores different square sizes and positions, pruning paths that exceed the current minimum solution.

Pruning to Optimize Search

Pruning is a key optimization that prevents unnecessary exploration. Whenever a solution exceeds the best-known solution, it is pruned to avoid redundant checks, making the search more efficient.

Memoization to Avoid Redundancy

To further optimize, memoization can be used to store intermediate results of subproblems. This prevents redundant calculations and speeds up the backtracking process by ensuring that previously computed states are not recalculated.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time and space complexity depend heavily on the final approach and the level of pruning applied. In the worst case, the time complexity can grow exponentially due to the backtracking nature, but pruning significantly reduces redundant paths, and memoization can further improve efficiency.

What Interviewers Usually Probe

  • Ability to implement backtracking efficiently
  • Understanding of pruning in search algorithms
  • Knowledge of dynamic programming optimizations like memoization

Common Pitfalls or Variants

Common pitfalls

  • Ignoring pruning, leading to unnecessary computation
  • Not using memoization, causing redundant calculations
  • Improperly handling edge cases like small rectangles

Follow-up variants

  • Allowing rectangles with larger dimensions
  • Implementing iterative solutions for large inputs
  • Exploring other tiling shapes or configurations beyond squares

FAQ

What is the key pattern for solving the Tiling a Rectangle with the Fewest Squares problem?

The key pattern is backtracking with pruning, where you explore all possible square sizes and placements while pruning unnecessary paths to minimize the number of squares.

How can I optimize my backtracking solution?

You can optimize by using pruning to eliminate paths that cannot lead to a solution better than the current one and applying memoization to avoid redundant calculations.

Can memoization help with large inputs?

Yes, memoization is especially helpful for large inputs as it stores intermediate results, reducing the number of redundant calculations during backtracking.

How do I handle edge cases in this problem?

Edge cases typically involve small rectangles, where no tiling is possible, or when the solution requires minimal squares. Ensure your algorithm can handle these scenarios by carefully managing the search space.

What other tiling problems are similar to this one?

Other tiling problems, such as tiling with different shapes or restricted tile sizes, also rely on backtracking and dynamic programming techniques to efficiently explore solution spaces.

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Solution

Solution 1: Recursive Backtracking + State Compression

We can perform recursive backtracking by position, during which we use a variable $t$ to record the current number of tiles used.

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class Solution:
    def tilingRectangle(self, n: int, m: int) -> int:
        def dfs(i: int, j: int, t: int):
            nonlocal ans
            if j == m:
                i += 1
                j = 0
            if i == n:
                ans = t
                return
            if filled[i] >> j & 1:
                dfs(i, j + 1, t)
            elif t + 1 < ans:
                r = c = 0
                for k in range(i, n):
                    if filled[k] >> j & 1:
                        break
                    r += 1
                for k in range(j, m):
                    if filled[i] >> k & 1:
                        break
                    c += 1
                mx = r if r < c else c
                for w in range(1, mx + 1):
                    for k in range(w):
                        filled[i + w - 1] |= 1 << (j + k)
                        filled[i + k] |= 1 << (j + w - 1)
                    dfs(i, j + w, t + 1)
                for x in range(i, i + mx):
                    for y in range(j, j + mx):
                        filled[x] ^= 1 << y

        ans = n * m
        filled = [0] * n
        dfs(0, 0, 0)
        return ans

Solution 2

#### Python3

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class Solution:
    def tilingRectangle(self, n: int, m: int) -> int:
        def dfs(i: int, j: int, t: int):
            nonlocal ans
            if j == m:
                i += 1
                j = 0
            if i == n:
                ans = t
                return
            if filled[i] >> j & 1:
                dfs(i, j + 1, t)
            elif t + 1 < ans:
                r = c = 0
                for k in range(i, n):
                    if filled[k] >> j & 1:
                        break
                    r += 1
                for k in range(j, m):
                    if filled[i] >> k & 1:
                        break
                    c += 1
                mx = r if r < c else c
                for w in range(1, mx + 1):
                    for k in range(w):
                        filled[i + w - 1] |= 1 << (j + k)
                        filled[i + k] |= 1 << (j + w - 1)
                    dfs(i, j + w, t + 1)
                for x in range(i, i + mx):
                    for y in range(j, j + mx):
                        filled[x] ^= 1 << y

        ans = n * m
        filled = [0] * n
        dfs(0, 0, 0)
        return ans
Tiling a Rectangle with the Fewest Squares Solution: Backtracking search with pruning | LeetCode #1240 Hard