<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:
        perfect_squares = []
        p = 1
        while p*p <=n:
            perfect_squares.append(p*p)
            p+=1
        
        num_count = 0
        q = deque()
        visited = set()
        
        q.append(n)
        visited.add(n)
        
        while q:
            length = len(q)
            for _ in range(length):
                front = q.popleft()
                if front == 0:
                    return num_count
                
                for ele in perfect_squares:
                    reduced_front = front - ele
                    if reduced_front >= 0 and reduced_front not in visited:
                        q.append(reduced_front)
                        visited.add(reduced_front)
            
            num_count +=1
        
        return num_count
        
17308