<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>

import numpy as np
class Solution:
    def numSquares(self, n: int) -> int:
        if n==1:
            return 1
        squares = [i*i for i in range(1,min(101, ceil(np.sqrt(n)))+1)]
        # print(squares)
        s = len(squares)
        if n in squares:
            return 1
        DP = [[n+1]*(n+1) for i in range(s+1)]

        for i in range(s+1):
            for j in range(n+1):
                if j==0:
                    DP[i][j] = 0
                elif i==0:
                    DP[i][j] = n+1
                else:
                    if squares[i-1]>j:
                        DP[i][j] = DP[i-1][j]
                    else:
                        DP[i][j] = min(DP[i-1][j], 1+DP[i][j-squares[i-1]])
        # print(DP)
        return DP[s][n]


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