<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>

from collections import defaultdict
class Solution:
    def numSquares(self, n: int) -> int:
        
        squares = []
        for i in range(1,n+1):
            if i*i <= n:
                squares.append(i*i)

        memo = defaultdict(lambda: float('inf'))
        def recursive(idx=0,rem=n):
            if rem == 0:
                return 0
            
            if rem < 0:
                return float('inf')

            if rem in memo:
                return memo[rem]
                
            for i in range(idx,len(squares)):
                if rem < squares[i]:
                    break
                memo[rem] = min(memo[rem],1 + recursive(i,rem-squares[i]))

            return memo[rem]

        return recursive()
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