<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:

    long long count(int n, int a, vector <vector <int>> &dp)
    {  
        if(n == 0)
        return 1;

        if(a < 1 && n != 0)
        return INT_MAX;

        if(dp[a][n] != -1)
        return dp[a][n];
        
        if(n >= a*a)
        return dp[a][n] = min(1 + count(n - a*a, a, dp), count(n, a-1, dp));

        else
        return dp[a][n] = count(n, a-1, dp);
    }
    
    int numSquares(int n) {

     int b = sqrt(n);
     if(b*b == n)
     return 1;

     int a = sqrt(n);

     vector <vector <int>> dp (a+1, vector <int> (n+1,-1));

    long long ans = count(n, a, dp);

    return (int)ans-1;

    }
};
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