<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:
        memo = {0: 0, 1: 1, 2: 2, 3: 3}
        def inner(n: int) -> int:
            if n in memo:
                return memo[n]
            square_num = 2
            min_seen = float(inf)
            squared_num = square_num**2
            while squared_num <= n: 
                cand_moves = inner(n - squared_num) + 1
                if cand_moves < min_seen: 
                    min_seen = cand_moves
                square_num += 1
                squared_num = square_num**2
            memo[n] = min_seen
            return min_seen
        return inner(n)

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