<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:
    def numSquares(self, n: int) -> int:
        # We want to try all possible numbers between 1 and max_num for every iteration
        # and reduce left sume by that value and increment count there
        # We look for smallest possible count

        # Use cache based on the left sum

        max_num = 1
        for num in range(1, n):
            if num ** 2 == n:
                return 1
            
            if num ** 2 > n:
                break
            max_num = num

        cache = {}
        
        def process_sum(left_sum):
            if left_sum < 0:
                return float('inf')
            
            if left_sum == 0:
                return 0
            
            if left_sum in cache:
                return cache[left_sum]
            
            min_count = float('inf')
            for num in range(1, max_num + 1):
                curr_count = 1 + process_sum(left_sum - (num ** 2))
                min_count = min(min_count, curr_count)
            
            cache[left_sum] = min_count
            return cache[left_sum]
        
        return process_sum(n)



            
            

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