How do you solve Tiling a Rectangle with the Fewest Squares on LeetCode?

Tiling a Rectangle with the Fewest Squares (LeetCode #1240) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m * min(n, m)) time complexity and O(n * m) space complexity. Key insight: Understanding how to break down the rectangle into smaller squares is crucial.

What is the brute force solution for Tiling a Rectangle with the Fewest Squares?

The brute force approach for Tiling a Rectangle with the Fewest Squares is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries all possible placements of squares in the rectangle, counting how many squares are needed to fill the area. This method is straightforward but can be inefficient as it explores many unnecessary configurations. The optimal Optimal Solution approach improves this to O(n * m * min(n, m)).

What algorithm or data structure is used to solve Tiling a Rectangle with the Fewest Squares?

Tiling a Rectangle with the Fewest Squares uses the following concepts: Backtracking. The recommended patterns to study are: Dynamic Programming, Backtracking.

What are the interview tips for Tiling a Rectangle with the Fewest Squares?

Explain your thought process clearly; interviewers appreciate understanding your approach. Don’t hesitate to ask clarifying questions about the problem constraints. Practice coding the solution on a whiteboard or in a shared document to simulate the interview environment.

What are common mistakes when solving Tiling a Rectangle with the Fewest Squares?

Not considering the base cases correctly, especially when n equals m. Failing to optimize recursive calls leading to excessive time complexity.

#1240

Tiling a Rectangle with the Fewest Squares

Hard

BacktrackingDynamic ProgrammingBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m * min(n, m))
Space	O(1)	O(n * m)

💡

Intuition

Time O(n * m * min(n, m))Space O(n * m)

The optimal solution uses dynamic programming to store the minimum number of squares needed for each sub-rectangle. This avoids redundant calculations and significantly speeds up the process.

⚙️

Algorithm

4 steps

1Step 1: Create a DP array where dp[i][j] represents the minimum number of squares needed for a rectangle of size i x j.
2Step 2: Initialize dp[i][i] = 1 for all i, as a square can fill a square rectangle.
3Step 3: For each rectangle size, iterate through all possible square placements, updating the DP array with the minimum count found.
4Step 4: Return dp[n][m] as the result.

solution.py10 lines

1def minSquares(n, m):
2    dp = [[float('inf')] * (m + 1) for _ in range(n + 1)]
3    for i in range(1, n + 1):
4        for j in range(1, m + 1):
5            if i == j:
6                dp[i][j] = 1
7            else:
8                for k in range(1, min(i, j) + 1):
9                    dp[i][j] = min(dp[i][j], 1 + dp[i - k][j], 1 + dp[i][j - k])
10    return dp[n][m]

ℹ

Complexity note: The time complexity is O(n * m * min(n, m)) because we fill a DP table of size n x m, and for each cell, we may iterate through the minimum dimension to find the best square placement.

1Understanding how to break down the rectangle into smaller squares is crucial.
2Dynamic programming helps avoid redundant calculations and speeds up the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.