<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 solve(int i, int n, vector<int>& sq, vector<vector<int>>& dp) {
        if (n == 0)
            return 0; 
        if (i >= sq.size() || n < 0)
            return INT_MAX;  
        if (dp[i][n] != -1)  return dp[i][n];

        int take = solve(i, n - sq[i], sq, dp);
        if (take != INT_MAX) take++; 
        int not_take = solve(i + 1, n, sq, dp);
        dp[i][n] = min(take, not_take);
        return dp[i][n];
    }

    int numSquares(int n) {
        if (ceil(sqrt(n)) == floor(sqrt(n))) {
            return 1; 
        }
        vector<int> sq;
        for (int i = 1; i <= (int)sqrt(n); i++) {
            sq.push_back(i * i); 
        }
        vector<vector<int>> dp(sq.size(), vector<int>(n + 1, -1));

        return solve(0, n, sq, dp);
    }
};
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