<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 helper(vector<int> &choices, int target) {
        int n = choices.size();
        vector<vector<int>> dp(n, vector<int>(target + 1, INT_MAX)); 
        
        for (int j = 0; j <= target; j++) {
            dp[0][j] = j;
        }

        for (int i = 1; i < n; i++) {
            for (int j = 0; j <= target; j++) {
                int notTake = dp[i-1][j]; 
                int take = INT_MAX;
                if (choices[i] <= j) {
                    take = dp[i][j - choices[i]] + 1;
                }
                dp[i][j] = min(take, notTake);
            }
        }
        return dp[n-1][target];
    }

    int numSquares(int n) {
        vector<int> choices;
        for (int i = 1; i * i <= n; i++) {
            choices.push_back(i * i);
        }
        return helper(choices, n);
    }
};

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