<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):

        # list of square numbers that are less than `n`
        square_nums = [i * i for i in range(1, int(n**0.5)+1)]
    
        level = 0
        queue = {n}
        while queue:
            level += 1
            #! Important: use set() instead of list() to eliminate the redundancy,
            # which would even provide a 5-times speedup, 200ms vs. 1000ms.
            next_queue = set()
            # construct the queue for the next level
            for remainder in queue:
                for square_num in square_nums:    
                    if remainder == square_num:
                        return level  # find the node!
                    elif remainder < square_num:
                        break
                    else:
                        next_queue.add(remainder - square_num)
            queue = next_queue
        return level
17308