<p>Given a rectangle of size <code>n</code> x <code>m</code>, return <em>the minimum number of integer-sided squares that tile the rectangle</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/17/sample_11_1592.png" style="width: 154px; height: 106px;" /></p>

<pre>
<strong>Input:</strong> n = 2, m = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> <code>3</code> squares are necessary to cover the rectangle.
<code>2</code> (squares of <code>1x1</code>)
<code>1</code> (square of <code>2x2</code>)</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/17/sample_22_1592.png" style="width: 224px; height: 126px;" /></p>

<pre>
<strong>Input:</strong> n = 5, m = 8
<strong>Output:</strong> 5
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/17/sample_33_1592.png" style="width: 224px; height: 189px;" /></p>

<pre>
<strong>Input:</strong> n = 11, m = 13
<strong>Output:</strong> 6
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n, m &lt;= 13</code></li>
</ul>

class Solution:
    def tilingRectangle(self, n: int, m: int) -> int:
        if n > m:
            n, m = m, n
        
        def dfs(heights, moves):
            if moves >= self.min_moves:
                return
            if all(h == n for h in heights):
                self.min_moves = min(self.min_moves, moves)
                return
            
            min_height = min(heights)
            start = heights.index(min_height)
            end = start
            while end < m and heights[end] == min_height:
                end += 1
            
            for size in range(min(end - start, n - min_height), 0, -1):
                new_heights = heights[:]
                for i in range(start, start + size):
                    new_heights[i] += size
                dfs(new_heights, moves + 1)
        
        self.min_moves = max(n, m)
        dfs([0] * m, 0)
        return self.min_moves
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