<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:
        def isPerfectSquare(x):
            if(x >= 0):
                sr = int(math.sqrt(x))
                return ((sr*sr) == x)
            return false
        q = deque()
        squares = []
        vis = set()
        for i in range(1,n +1):
            if isPerfectSquare(i):
                squares.append(i)
        q.append((0, 1))
        vis.add((0,1))
        while q:
            curr = q.popleft()
            curr_sum, count = curr
            for i in range(len(squares)):
                new_sum = curr_sum + squares[i]
                if new_sum == n:
                    return count
                
                if (new_sum, count+1) in vis:
                    continue
                q.append((new_sum, count+1))
                vis.add(((new_sum, count+1)))

                


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