<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:
        # dp = [float('inf')] * (n + 1)
        # dp[0] = 0
        # dp[1] = 1
        # for i in range(1, n + 1):
        #     min_val = float('inf')
        #     j = 1
        #     while j * j <= i:
        #         min_val = min(min_val, dp[i - j * j] + 1)
        #         j += 1
        #     dp[i] = min_val
        # return dp[n]

        @cache
        def helper(i: int):
            min_cnt = math.inf

            if i == 0:
                return 0
            # if i == 1:
            #     return 1
            for j in range(1, i + 1):
                if j * j <= i:
                    cur_cnt = helper(i - j * j) + 1
                    min_cnt = min(min_cnt, cur_cnt)
                else:
                    break
            return min_cnt
        
        return helper(n)
32631