<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:
        
        memo = {}

        def dfs(rem):
            if rem == 0:
                return 0

            if rem in memo:
                return memo[rem]

            i = 1
            min_count = float('inf')

            while i * i <= rem:
                min_count = min(min_count,dfs(rem - i*i)+1)
                memo[rem] = min_count
                i += 1
            return memo[rem]

        return dfs(n)


        # #tc = o(n.n^1/2)
        # dp = [n] * (n+1)
        # dp[0] = 0
        
        # for target in range(1,n+1):
        #     for s in range(1,target+1):
        #         square = s * s
        #         if target - square < 0:
        #             break
        #         dp[target] = min(dp[target],1+dp[target-square])
        # return dp[n]
31884