<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:
        # dp
#         dp = [float("inf")] * (n + 1)
#         dp[0] = 0

#         for i in range(1, n + 1):
#             j = 1
#             while j * j <= i:
#                 dp[i] = min(dp[i], dp[i - j * j] + 1)
#                 j += 1

#         return dp[n]

        #BFS
        queue = [n]
        visited = set([n])
        step = 0
        while len(queue) > 0:
            step += 1
            for _ in range(len(queue)):
                curr = queue.pop(0)
                for i in range(1, int(math.sqrt(n)) + 1):
                    next_val = curr - i * i
                    if next_val == 0:
                        return step
                    if next_val not in visited:
                        queue.append(next_val)
                        visited.add(next_val)
                        
        return step
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