<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:
        cache = [-1] * (n+1)
        
        def recursion(val):
            if cache[val] != -1:
                return cache[val]
            
            if val == n:
                return 0
            if val > n:
                return float("inf")
            
            min_tries = float("inf")
            for i in range(1, int(math.sqrt(n)) + 1):
                square = i * i
                if val + square <= n:
                    min_tries = min(min_tries, 1 + recursion(val + square))
            
            cache[val] = min_tries
            return min_tries
        
        return recursion(0)
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