<p>Given an integer <code>n</code>, return <em>the least number of perfect square numbers that sum to</em> <code>n</code>.</p>

<p>A <strong>perfect square</strong> is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, <code>1</code>, <code>4</code>, <code>9</code>, and <code>16</code> are perfect squares while <code>3</code> and <code>11</code> are not.</p>

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

<pre>
<strong>Input:</strong> n = 12
<strong>Output:</strong> 3
<strong>Explanation:</strong> 12 = 4 + 4 + 4.
</pre>

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

<pre>
<strong>Input:</strong> n = 13
<strong>Output:</strong> 2
<strong>Explanation:</strong> 13 = 4 + 9.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def numSquares(self, n: int) -> int:
        x=[]
        if n<=1:
            return n
        for i in range(1,n):
            m=i*i
            x.append(m)
            if m>n:
                x.pop()
                break
        m=len(x)
        dp=[[9999999 for i in range(n+1)] for j in range(m)]
        m=len(x)
        for i in range(m):
            dp[i][0]=0
        for i in range(m):
            for j in range(1,n+1):
                if j>=x[i]:
                    dp[i][j]=min(dp[i-1][j],1+dp[i][j-x[i]])
                else:
                    dp[i][j]=dp[i-1][j]
        return dp[-1][-1]
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