<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:
        # Generate all perfect squares less than or equal to n
        arr = []
        i = 1
        while i * i <= n:
            arr.append(i * i)
            i += 1

        m = len(arr)
        # Initialize the DP table
        t = [[float('inf')] * (m + 1) for _ in range(n + 1)]

        # Base case: 0 can be made with 0 numbers
        for j in range(m + 1):
            t[0][j] = 0

        # Fill the DP table
        for i in range(1, n + 1):
            for j in range(1, m + 1):
                if arr[j - 1] <= i:
                    t[i][j] = min(t[i][j-1], 1 + t[i - arr[j - 1]][j])
                else:
                    t[i][j] = t[i][j-1]

        # The answer will be in t[n][m], which gives the minimum number of squares that sum to n
        return t[n][m]
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