<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> squares;
        
        for (int i = 1; i*i <= n; ++i) {
            int square = i*i;
            
            if (square == n) {
                return 1;
            }
            
            squares.push_back(square);
        }
        
        queue<pair<int, int>> q;
        set<int> s;
        
        q.push({0, 0});
        
        while (!q.empty()) {
            auto [curSum, curCnt] = q.front();
            q.pop();
            
            for (int square : squares) {
                int newSum = curSum + square;
                
                if (newSum == n) {
                    return curCnt + 1;
                }
                
                if (newSum > n) {
                    break;
                }
                
                else {
                    q.push({newSum, curCnt + 1});
                }
                
            }
            
        }
        
        return -1;
    }
};
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