<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) {
        for (int i = 1; i * i <= n; ++i) {
            squares.insert(i * i);
        }

        dp.push({n, 0});

        return bfs();
    }
private:
    set<int> squares;
    queue<pair<int, int>> dp;

    int bfs() {
        while (!dp.empty())
        {
            auto x = dp.front();
            dp.pop();

            if (squares.contains(x.first)) {
                return x.second + 1;
            }
            
            for (auto s : squares) {
                if (x.first - s > 0) {
                    dp.push({x.first - s, x.second  + 1});
                }
            }
        }
        return -1;
    }
};
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