<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>

from collections import deque
import math
class Solution:
    def numSquares(self, n: int) -> int:
        queue = deque([(0, n)])
        cur_min = float('inf')
        seen = set()
        while queue:
            cur_count, target = queue.popleft()
            if cur_count >= cur_min and target != 0:
                continue
            if target == 1 or (target ** 0.5) % 1 == 0:
                cur_min = min(cur_min, cur_count + 1)
                continue
            root = math.ceil(target ** 0.5)
            for i in range(1, root):
                t = target - i ** 2
                if t == 0:
                    cur_min = min(cur_min, cur_count + 1)
                else:
                    to_add = (cur_count + 1, t)
                    if to_add not in seen:
                        queue.append((cur_count + 1, t))
                        seen.add((cur_count + 1, t))
        return cur_min
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