<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 {
public:     
    int dp[201][10001];
    int rec(int ind,int target,vector<int>&a,int nn){
        if(ind==nn-1){
            if(target%a[nn-1]==0)return target/a[nn-1];
            else return 1e9;
        }
        if(dp[ind][target]!=-1)return dp[ind][target];
        int ntake=rec(ind+1,target,a,nn);
        int take=1e9;
        if(target>=a[ind])take=1+rec(ind,target-a[ind],a,nn);
        return dp[ind][target]=min(take,ntake);
    }
    int numSquares(int n) {
        int r=sqrt(n);
        vector<int>a;
        memset(dp,-1,sizeof(dp));
        for(int i=1;i<=r;i++)a.push_back(i*i);
        int nn=a.size();
        return rec(0,n,a,nn);

    }
};
18180