<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)
    {
        cache[0] = INT_MAX;
        check(n, sqrt(n), 0);
        return cache[0];
    }

private:
    unordered_map<int, int> cache;

    bool check(int num, int square, int depth)
    {
        if (!num)
        {
            cache[0] = min(cache[0], depth);
            return true;
        }

        if (cache.contains(num) && cache[num] <= depth) return false;
        else cache[num] = depth;

        while (square * square > num) square--;

        while (square) 
        {
            check(num - square * square, square, depth + 1);
            square--;
        }

        return true;
    }
};
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