<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) {
        std::unordered_set<int> visited;
        std::queue<int> red;
        
        int duzina = std::sqrt(n);
        std::vector<int> neighbors(duzina);
        for(int i=1; i<=duzina; i++) {
            neighbors[i - 1] = i * i;    
        }
        
        int steps{};
        red.push(0);
        
        while(!red.empty()) {
            int len = red.size();
            for(int i=0; i<len; i++) {
                int current = red.front();
                if(current == n) {
                    return steps;
                }
                for(int next : neighbors) {
                    int tmp = current + next;
                    if(visited.count(tmp) == 0) {
                        visited.insert(tmp);
                        red.push(tmp);
                    }
                }
                red.pop();
            }
            steps++;
        }
        
        return n;
    }
};
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