<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:
        # f(n) = f(n-1) + f(n-4)
        self.cache = {}
        return self.helper(n)
    
    def helper(self, n):
        if n == 0:
            return 0
        if n in self.cache:
            return self.cache[n]
        # cal min
        min_val = float("inf")
        max_n = self.find_max(n)
        for i in range(1, max_n + 1):
            min_val = min(min_val, self.helper(n - i * i))
        self.cache[n] = min_val + 1
        return min_val + 1
    def find_max(self, n):
        for i in range(1, n + 1):
            if i * i > n:
                return i - 1
        return n

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