Introduction

In this blog post, we will explore how to solve the LeetCode problem "1240" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a rectangle of size n x m, return the minimum number of integer-sided squares that tile the rectangle. Example 1: Input: n = 2, m = 3 Output: 3 Explanation: 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 3: Input: n = 11, m = 13 Output: 6 Constraints: 1 <= n, m <= 13

Explanation

Here's the approach, complexity analysis, and Python code:

• Key Idea: Utilize dynamic programming (DP) to store and reuse results for sub-rectangles. The minimum number of squares needed to tile a rectangle (n, m) can be found by trying all possible cuts (horizontal or vertical) and taking the minimum of the sums of the squares needed for the resulting sub-rectangles.
• Memoization: Use a dictionary (or a 2D array) to store the results of already computed sub-rectangles to avoid redundant calculations.

• Base Case: If n == m, then only one square is needed.

• Complexity: O(n*m*min(n, m)) runtime, O(n*m) storage

Code

    def tilingRectangle(n: int, m: int) -> int:
    """
    Given a rectangle of size n x m, return the minimum number of integer-sided squares that tile the rectangle.

    Example 1:
    Input: n = 2, m = 3
    Output: 3
    Explanation:
    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 3:
    Input: n = 11, m = 13
    Output: 6

    Constraints:
    1 <= n, m <= 13
    """

    dp = {}  # Memoization table

    def solve(n, m):
        if n == m:
            return 1
        if (n, m) in dp:
            return dp[(n, m)]
        if (m, n) in dp:
            return dp[(m, n)]

        ans = float('inf')

        # Horizontal cuts
        for i in range(1, n):
            ans = min(ans, solve(i, m) + solve(n - i, m))

        # Vertical cuts
        for i in range(1, m):
            ans = min(ans, solve(n, i) + solve(n, m - i))

        dp[(n, m)] = ans
        return ans

    return solve(n, m)