<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:
        memo = {} 
        all_squares = {} 
        for i in range(1, n):
            if i * i > n:
                break 
            all_squares[i] = i * i 

        def dp(i, memo, all_squares):
            # returns the least number of perfect square numbers
            if i in memo:
                return memo[i]

            # bc: given that our starting number is 0, we know that it will have no perfect squares?
            if i < 4:
                return i 
            
            all_poss_squares = [] 
            for key in all_squares:
                if all_squares[key] <= i:
                    all_poss_squares.append(all_squares[key])
             

            total = float('inf')
            for square in all_poss_squares:
                total = min(total, dp(i-square, memo, all_squares) + 1)
            
            memo[i] = total
            return total 
        
        return dp(n, memo, all_squares)

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