<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>

import math

class Solution:
    def numSquares(self, n: int) -> int:
        squares = [x ** 2 for x in range(1, n + 1)]
        dp = {}

        def helper(num):
            res = float('inf')
            if num < 0: return float('inf')
            if num == 0: return 0
            if num not in dp:
                i = 0
                while i < len(squares) and squares[i] <= num: i += 1
                for x in range(i - 1, -1, -1):
                    if x > num: break
                    res = min(res, helper(num - squares[x]) + 1)
                dp[num] = res
            return dp[num]
                
        return helper(n)
41601