<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) {
        int t = sqrt(n);
        vector<int> coins(t,0);
        for(int i=1;i<=t;i++){
            coins[i-1]=i*i;
        }
        vector<vector<int>> dp(t,vector<int>(n+1,0));
        for(int i=1;i<=n;i++){
            if(i%coins[0]==0){
                dp[0][i]=i/coins[0];
            }else{
                dp[0][i]=1e9;
            }
        }
        for(int i=1;i<t;i++){
            for(int j=1;j<=n;j++){
                int nt = dp[i-1][j];
                int tk = 1e9;
                if(coins[i]<=j){
                    tk = 1+dp[i][j-coins[i]];
                }
                dp[i][j]=min(tk,nt);
            }
        }
        return dp[t-1][n];

    }
};
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