<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>

'''
Notes: DP

'''
class Solution:
    def numSquares(self, n: int) -> int:
        def dfs(current_sum, step):
            #print(f'dfs: current_sum: {current_sum}, step: {step}')
            nonlocal least_number

            if current_sum in cache:
                return cache[current_sum] + step

            if current_sum > n:
                return n
            
            if current_sum == n:
                return step

            remain = n - current_sum
            i = 1
            square = 1

            while remain - square >= 0:
                least_number = min(least_number, dfs(current_sum + square, step+1))
                i += 1
                square = i * i

            cache[current_sum] = least_number - step
            return least_number

        cache = {}
        least_number = n
        return dfs(0,0)
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