<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:

        # first we need to get perfect square numbers lists
        nums = [0]
        i = 1
        while i ** 2 <= n:
            nums.append(i ** 2)
            i += 1

        row, column = len(nums), n+1
        # dp[i][j] = least number of perfect square numbers 
        # that sum to j with first i+1 perfect square numbers
        dp = [[float('inf')] * column for _ in range(row)]

        # initialze: when target is 0, need 0 numbers
        for i in range(row):
            dp[i][0] = 0
        
        # start:
        for i in range(1, row): # curr_num = nums[i]
            for j in range(1, column): # curr_target = j
                curr_num = nums[i]
                if j >= curr_num:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-curr_num]+1)
                else: 
                    dp[i][j] = dp[i-1][j]
        
        return dp[-1][n]


        
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