<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:
        if (n ** (1/2)) % 1 == 0:
            return 1
        squares=[i**2 for i in range(int(n**(1/2)),0,-1)]
        l=len(squares)
        counter = 2
        queue = deque((squares[i], i) for i in range(len(squares) - 1))
        print(queue)
        while queue:
            for _ in range(len(queue)):
                s = queue.popleft()
                for i in range(s[1], len(squares)):
                    new_s = s[0] + squares[i]
                    if new_s == n:
                        return counter
                    if new_s < n:
                        queue.append((new_s, i))
            counter += 1
            print(queue)
        return n
27399