<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>

from functools import lru_cache
from math import ceil
class Solution:
    def numSquares(self, n: int) -> int:
        
        squares = [i*i for i in range(1, ceil(n**0.5)+1)]
        squares = squares[::-1]
        ans = []

        s = n
        n = len(squares)

        dp = [[10000000 for _ in range(s+1)] for _ in range(n+1)]

        dp[0][0] = 0
        for i in range(1, n+1):
            dp[i][0] = 0

        for i in range(1, n+1):

            for j in range(1, s+1):

                dp[i][j] = dp[i-1][j]
                if j >= squares[i-1]:
                    dp[i][j] = min(dp[i][j], dp[i][j-squares[i-1]]+1)
        
        return dp[n][s]
24783