<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:
        
        # from int n, we have the option of subtracting all perfect squares that are less than n 
        # first we can generate all perfect squares less than n
        def generate_squares(n):
            ret = []
            i = 1
            while i**2 <= n:
                ret.append(i**2)
                i += 1
            return ret

        # we will check if n - perfect square equals 0, if so we return the jumps it took
        queue = collections.deque()
        queue.append(n)
        visited = set()
        steps = 1
        while queue:
            curr_step = len(queue)
            for _ in range(curr_step):
                curr_n = queue.popleft()
                # generate all paths for curr_n and append to queue
                if curr_n in visited:
                    continue
                paths = generate_squares(curr_n)
                visited.add(curr_n)
                for path in paths:
                    next_n = curr_n - path
                    if next_n == 0:
                        return steps
                    queue.append(next_n)
                    
            steps += 1

        return -1
        # we can also track numbers we have visisted that have not led to solutions and avoid them
27773