<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 func(int ind, vector<int> &pfsq, int n,vector<vector<int>> &dp){
        if(ind < 0) return 1e9;


        if(dp[ind][n]!=-1) return dp[ind][n];

        if(n == 0) return 0;
        int t=1e9,nt=1e9;

        if(pfsq[ind]<= n){
            t = 1+func(ind, pfsq, n-pfsq[ind],dp);
        }

        nt = func(ind-1,pfsq,n,dp);

        return dp[ind][n] = min(t,nt);
    }

    int numSquares(int n) {
        vector<int> pfsq;
        int i=1;
        while(i*i<=n){
            pfsq.push_back(i*i);
            i++;
        }

         vector<vector<int>> dp(pfsq.size(),vector<int> (n+10,-1));

        return func(pfsq.size()-1,pfsq,n,dp);
    }
};
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