<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:
        squares = []
        num = 1
        while True:
            num_square = num**2
            if num_square>n:
                break
            squares.append(num_square)
            num += 1
        
        dp = [[0]*len(squares) for _ in range(n+1)]
        
        for i in range(1,len(dp)):
            for j in range(len(dp[0])):
                if i - squares[j]>=0:
                    if j == 0:
                        dp[i][j] = 1+ dp[i-squares[j]][j]
                    else:
                        dp[i][j] = min(dp[i][j-1], 1+dp[i-squares[j]][j])
                else:
                    dp[i][j] = dp[i][j-1]
        
        #print(dp)
        return dp[-1][-1]

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