<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 cnt, vector<int> &v, int rem,
    vector<vector<int>> &dp)
    {
        if(ind==0)
        {
            if(rem%v[0]==0)
                return rem/v[0];
            return INT_MAX;
        }
        if(rem==0)
            return cnt;
        if(dp[ind][rem]!=-1)
            return dp[ind][rem];
        int notPick = f(ind-1, cnt, v, rem, dp);
        int pick = INT_MAX;
        if(v[ind]<=rem)
            pick = 1+f(ind, cnt, v, rem-v[ind], dp);
        return dp[ind][rem] = min(pick, notPick);
    }
    int numSquares(int n) 
    {
        vector<int> v;
        for(int i=1;i*i<=n;i++)
            v.push_back(i*i);
        int cnt=0;
        int size = v.size();
        vector<vector<int>> dp(size, vector<int>(n+1, -1));
        cnt = f(size-1, cnt, v, n, dp);
        return cnt;
    }
};
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