<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:
    # n, sum, 
    # we can either apply 1, 2, 4, 9.. square sum
    def smaller_perfect_squares(self, target):
        arr = list()
        for i in range(1, target+2):
            if i*i > target:
                return arr
            arr.append(i*i)
    
    # goal of this function, is, given n, it will return the least number ....
    def numSquaresAux(self, n, cache):
        if n in cache:
            return cache[n]
        
        if n == 0:
            return 0
        
        candidates = self.smaller_perfect_squares(n)
        
        best_count = float("inf")
        for can in candidates:
            if n-can < 0:
                continue
                
            can_count = self.numSquaresAux(n-can, cache)
            
            if (can_count +1) < best_count:
                best_count = (can_count +1)
        cache[n] = best_count
        return best_count
        
    
    
    def numSquares(self, n: int) -> int:
        cache = dict()
        return self.numSquaresAux(n, cache)
        
        
        
41228