<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:
        self.ans=float('inf')
        def backtrack(height, moves):
            if min(height)==n:
                self.ans=moves
            
            if moves>=self.ans:return
            i=min_height_idx=height.index(min(height))

            while i< len(height) and height[i]==height[min_height_idx]:
                i+=1
            
            mx_sq_side=min(n-height[min_height_idx] , i-min_height_idx)

            for h in range(1,mx_sq_side+1):
                new_height = height[:]
                for j in range(min_height_idx,min_height_idx+h):
                    new_height[j]+=h
                backtrack(new_height,moves+1)
        backtrack([0]*m,0)
        return self.ans
16400