<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 math
class Solution(object):
    def new_subset(self,i,arr,target,dp):
        if(target==0):
            return 0
        if(i<0):
            return float('inf')
        if(dp[i][target]!=-1):
            return dp[i][target]
        notTake=0+self.new_subset(i-1,arr,target,dp)
        Take=float('inf')
        if(arr[i]<=target):
            Take=1+self.new_subset(i,arr,target-arr[i],dp)
        dp[i][target]=min(Take,notTake)
        return dp[i][target]
    def numSquares(self, n):
        """
        :type n: int
        :rtype: int
        """
        sqrt=int(math.sqrt(n))
        new_list=[]
        for i in range(1,sqrt+1):
            new_list.append(i*i)
        ind=len(new_list)
        dp=[[-1 for i in range(n+1)]for j in range(ind)]
        t=self.new_subset(ind-1,new_list,n,dp)
        return t
27399