<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 {
    private:
    bool isPerfect(int i) {
    int root = sqrt(i);
    return root * root == i;
}

public:
    int numSquares(int n) {
        // if(n==1 ) return 1;
        vector<int>perfect;
        for(int i=1;i<=n;i++){
            if(isPerfect(i)){
                perfect.push_back(i);
            }
        }
        int s=perfect.size();
       // coin change tep do
       vector<vector<int>>dp(s,vector<int>(n+1,0));
       for(int i=1;i<=n;i++){
          if(i%perfect[0]==0) dp[0][i]=i/perfect[0];
          else dp[0][i]=1e9;
       }
       for(int i=1;i<s;i++){
        for(int j=0;j<=n;j++){
            int notake=0+dp[i-1][j];
            int take=1e9;
            if(perfect[i]<=j)  take=1+dp[i][j-perfect[i]];
            dp[i][j]=min(take,notake);
        }
        
       }
         int ans=dp[s-1][n];
         return ans;
        // return perfect[3];
    }
};
154766