<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:
        if n ==1 : return 1

        

        # dp = [n]* (n+1)
        # dp[0] = 0
        # # print(dp)

        # for num in range(1,n+1): 
        #     for i in range(1,num+1):
        #         s = i*i
        #         if num - s < 0: 
        #             break
        #         dp[num] = min(dp[num], 1 + dp[num-s])
     
        # return dp[n]
        memo = {}

        def dfs(x: int) -> int:
            if x in memo:
                return memo[x]
            if x == 0:
                return 0
            min_count = (x+1)*(x+1)
            i = 1
            while i * i <= x:
                min_count = min(min_count, 1 + dfs(x - i * i))
                i += 1
            memo[x] = min_count
            return min_count

        return dfs(n)






        
43096