<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>

#define LARGE 9999
class Solution {
public:
    int numSquares(int n) {
        vector<int> mem (n+1, LARGE);
        vector<int> nextMem (n+1, LARGE);
        for (int i = 0; i < mem.size(); i++){
            mem[i] = i;
        }
        for (int level = 2; level * level <= n; level++){
            const int addition = level * level;
            for (int i = 0; i < mem.size(); i++){
                const int prevMemValue = i - addition;
                int bestCost = mem[i];
                if (prevMemValue >= 0){
                    bestCost = min(bestCost, 1 + nextMem[prevMemValue]);
                    bestCost = min(bestCost, 1 + mem[prevMemValue]);
                }
                nextMem[i] = bestCost;
            }
            swap(mem, nextMem);
            nextMem = vector<int> (n+1, LARGE);
        }
        return (mem.back() >= LARGE) ? -1 : mem.back();
    }
};
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