<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:


        max_n = int(sqrt(n))

        search_range = [i ** 2 for i in range(1, max_n + 1)]
        '''
        dp = [float('inf')] * (n + 1)
        dp[0] = 0 

        for i in range(len(search_range)):
            for j in range(search_range[i], len(dp)):
                dp[j] = min(dp[j], 1 + dp[j - search_range[i]])

        return dp[-1]'''




        def helper(i, total):
            
            if total == n:
                return 0 
            
            if total in memo:
                return memo[total]
            
            memo[total] = float('inf')

            for j in range(i, len(search_range)):
                if total + search_range[j] <= n:                  
                    memo[total] = min(memo[total], 1 + helper(j, total + search_range[j]))

            return memo[total]

        memo = {}
        result = float('inf')
        for i in range(len(search_range)):
            result = min(result, helper(i, 0))

        return result




        
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