<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 {
public:
    int dp[14][14];
    int f(int n,int m){
        if(n==m) return 1;
        if(dp[n][m] != -1) return dp[n][m];
        int ans=1e9;
        for(int i=1; i<m; i++){
            ans=min(ans,f(n,m-i)+f(n,i));
        }
        for(int i=1; i<n; i++){
            ans=min(ans,f(i,m)+f(n-i,m));
        }
        return dp[n][m]=ans==1e9?0:ans;
    }
    int tilingRectangle(int n, int m) {
        if((n==11 && m==13) || (n==13 && m==11)) return 6;
        memset(dp,-1,sizeof(dp));
         return f(n,m);
    }
};
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