<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(object):
    def numSquares(self, n):
        
        # Initialize a list to store the minimum number of perfect squares required for each number up to n
        dp = [0] * (n + 1)
         
        for i in range(1, n + 1):
            # Initialize the value for dp[i] to i, as it's the worst case (when n is the sum of i 1s)
            dp[i] = i
            j = 1
            # Go through the perfect squares: 1, 4, 9, etc... for each j
            while j * j <= i:
                # For each square number j * j less than or equal to i, update dp[i] if needed
                # Below we check the complement of i with the square (add 1 because we have the square as well)
                # to see if less than current
                dp[i] = min(dp[i], dp[i - j * j] + 1) # we use the prior via or dp (dynamic programming)
                j += 1
         
        return dp[n]
16186