<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:
        # if n is perfect square, return 1
        # else
        # break down to smaller problems. Take 1 + min of
        # f(n - 1), f(n - 4), f(n - 9) ... f(1)
        # cache everything

        cache = [0 for _ in range(n + 1)]
        p_squares = [x * x for x in range(1, math.ceil(math.sqrt(n)) + 1)]
        for square in p_squares:
            if square < n + 1:
                cache[square] = 1

        def f(x: int) -> int:
            if cache[x] > 0:
                return cache[x]

            result = 1 + min(
                [f(x - k) for k in p_squares if x > k],
                default=0,
            )
            cache[x] = result
            return result
        return f(n)

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