<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:

    squares = []

    memo = dict()

    def __init__(self):
        i = 1
        while i*i <= 10**4:
            self.squares.append(i*i)
            i+=1

    def recNumSquares( self, index, n ):

        if n == 0:
            return 0

        if index >= len(self.squares) or self.squares[index] > n:
            return 1000000

        if n < 0:
            return 1000000

        if n in self.memo:
            return self.memo[n]

        # we can either either pick this index again or move to the next
        pickedVal = 1 + self.recNumSquares(index, n - self.squares[index] )

        notPickedVal = self.recNumSquares(index + 1,  n)

        self.memo[n] = min(pickedVal, notPickedVal)

        return min(pickedVal, notPickedVal)

    def numSquares(self, n: int) -> int:

        return self.recNumSquares(0, n)



        
        
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