<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>

import numpy as np
class Solution:
    def numSquares(self, n: int) -> int:
        m = int(np.floor(np.sqrt(n)))
        nums = [10000] * (n + 1)  # nums[i] stores least nums of squares to get i
        nums[0] = 0
        for i in range(1, n + 1):
            # min of nums[i - squares[j]] + 1
            # nums = [0, 1, 2, 3, 1, 2]
            l = []
            for j in range(m + 1):
                if i - j ** 2 < 0:
                    break
                l.append(nums[i - j ** 2] + 1)
            nums[i] = min(l)
        print(nums)
        return nums[-1]
28894