<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 numSquares(int n) {
        vector<int>squares;
        vector<int>dp(n+1,n+1);
        dp[0]=0;
        if(n==1)return 1;
        for(int i=1;i<n;i++)
        {
            if(i*i>n)
            break;

            squares.push_back(i*i);
        }
        for(int i=squares[0];i<=n;i++)
        {
            for(int j=0;j<squares.size();j++)
            {
                if(i-squares[j]>=0)
                {
                    dp[i]=min(dp[i],dp[i-squares[j]]+1);
                }
            }
        }
        return dp[n];
        
    }
};
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