<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 f(int i, int sum, int tar, vector<int>&a, vector<vector<int>>&dp){
        if(sum == tar) return 0;
        if(sum > tar) return 1e9;
        if(i == a.size()) return 1e9;

        if(dp[i][sum] != -1) return dp[i][sum];
        int tk = 1 + f(i, sum+a[i], tar, a, dp);
        int ntk = f(i+1, sum, tar, a, dp);

        return dp[i][sum] =  min(tk, ntk);
        

    }

    int numSquares(int n) {
        vector<int>a;
        for(int i = 1; i*i <= n; i++){
            a.push_back(i*i);
        }
        int s = a.size();
        reverse(a.begin(), a.end());
        
        if(a[0] == n) return 1;

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

        return f(0, 0, n, a, dp);

    }
};
131681