<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 f(int ind,int total,vector<int>&v,vector<vector<int>>&dp)
        {
        if(ind==0)
            {
            if(total%v[ind]==0)
                return total/v[ind];
            return INT_MAX-1;
            }
        if(dp[ind][total]!=-1)
            return dp[ind][total];
        int nottake=f(ind-1,total,v,dp);
        int take=INT_MAX;
        if(total>=v[ind])
            take=1+f(ind,total-v[ind],v,dp);
        return dp[ind][total]=min(take,nottake);
        }
        
    int numSquares(int n) {
        vector<int>v;
        for(int i=1;i<=n;i++)
            {
            if(i*i==n)
                return 1;
            if(i*i<n)
            v.push_back(i*i);
            }
        for(auto i:v)
            {
            cout<<i<<endl;
            }
        vector<vector<int>>dp(v.size(),vector<int>(n+1,-1));
        return f(v.size()-1,n,v,dp);
    }
};
120139