<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>

"""

178
-169 + 9

the greedy solution is also optimal??
12 - 9 = 3
1 +1 +1 
NO


to find the sequence of perfect squares backwards from n we have to:
while n >= 1:
    new perfect square = int(sqrt(n))
    n = new_perfectsquare - 1

We have to explore all the possibilities:
 starting from n we start from the nearest perfect square -> int(sqrt(n))
 nps = int(sqrt(n))
 
the minimum number of ps from n is equal to the min(1 + min number(n - next ps))
if n==0 then is 0
i n ==1  then is 1
otherwise is the minimum number among choosing the other options

"""





class Solution:
    
    def numSquares(self, n: int) -> int:
        PERFECT_SQUARES = set([i*i for i in range(1, 100)])
        @cache
        def minNumberPerfectSquare(n: int) -> int:
            if n == 0:
                return 0
            if n in PERFECT_SQUARES:
                return 1
            
            min_num = n
            curr_p_root = int(n**(0.5))
            for root in range(curr_p_root, 1, -1):
                PERFECT_SQUARES.add(root*root)
                res = 1 + minNumberPerfectSquare(n - root*root)
                if res < min_num:
                    min_num = res
            return min_num
        
        return minNumberPerfectSquare(n)
                
    
    
    
    
    
    
    
    
    
        
36369