<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) {
        if(n==1)
            return 1;
        vector<int>a;
        int i=1,mini=20000,j;
        while(i*i<n){
            a.push_back(i*i);
            i++;
        }
        if(i*i==n)
            return 1;
        vector<vector<int>>dp(a.size()+1,vector<int>(n+1,0));
        for(i=0;i<dp[0].size();i++)
            dp[0][i]=INT_MAX;
        for(i=1;i<dp.size();i++){
            for(j=1;j<dp[0].size();j++){
                if(a[i-1]>j)
                    dp[i][j]=dp[i-1][j];
                else
                    dp[i][j]=min(dp[i][j-a[i-1]]+1,dp[i-1][j]);
            }
        }
        return dp[a.size()][n];
    }
};
129758