<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) {
        vector<int> results (n + 1, -1);
        int num = 1;
        vector<int> squares;
        set<int> seen;
        queue<pair<int, int>> visit;

        while (num * num <= n) {
            results[num * num] = 1;
            squares.push_back(num * num);
            num++;
        }
        if (squares.back() == n) {
            return 1;
        }

        for (int i = squares.size() - 1; i >= 0; i--) {
            visit.push({squares[i], 1});
        }

        while (!visit.empty()) {
            auto curr = visit.front();
            visit.pop();
            for (int square : squares) {
                int newVal = curr.first + square;
                if (newVal == n) {
                    return curr.second + 1;
                } else if (newVal < n) {
                    if (!seen.contains(newVal)) {
                        visit.push({newVal, curr.second + 1});
                        seen.insert(newVal);
                    }
                } else {
                    break;
                }
            }
        }
        return -1;
    }
};
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