<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 solve(int n){
        vector<bool> visit(100005, false);
        //vector<pair<int, int>> visit(100005);
        vector<int> squares;
        for(int i=1; i<=100; i++) squares.push_back(i*i);

        queue<pair<int, int>> bfs;
        for(int i=0; i<100; i++){
            if(squares[i]<=n) bfs.push({squares[i], 1});
        }
        
        while(!bfs.empty()){
            pair<int, int> current = bfs.front();
            int value = current.first;
            int level = current.second;
        
            bfs.pop();
            if(visit[value]) continue;
            visit[value] = true;
            
            if(value == n) return level;
            for(int i=0; i<100; i++){
                int aux = value+squares[i];
                if(aux > n) continue;
                bfs.push({value+squares[i], level+1});
            }

        }
        return 0;
    }
    int numSquares(int n) {
        return solve(n);
    }
};
108596