<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 sys
class Solution:
    def numSquares(self, n: int) -> int:

        
         end = int(n ** 0.5)
         perf = [i**2 for i in range(1, end + 1)]
      
    # Initialize dp array where dp[i] represents the minimum number of perfect squares that sum up to i
         dp = [float('inf')] * (n + 1)
         dp[0] = 0  # Base case: 0 can be formed with 0 perfect squares

         for i in range(1, n+1):
            for j in perf:
                if(j<=i):
                    dp[i] = min(dp[i], dp[i-j]+1)
    
    
        
         return dp[-1]
16560