<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:
        squares = []
        for num in range(1, isqrt(n)+1):
            squares.append(num**2)

        slen = len(squares)
        
        dp = [[0]*slen for i in range(n)]

        for i in range(1, n+1):
            for j in range(1, slen+1):
                if squares[j-1] <= i:
                    if i % squares[j-1] == 0:
                        dp[i-1][j-1] = i // squares[j-1]
                    else:
                        dp[i-1][j-1] = min(i // squares[j-1] + dp[i%squares[j-1]-1][j-2], dp[i-1][j-2])
                else:
                    dp[i-1][j-1] = dp[i-1][j-2]
        
        return dp[n-1][slen-1]
26651