<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){
        queue<pair<int,int>>  bfs;
        vector<bool> vist(10000, false);
        vector<int> squares;
        for(int i=1; i<=100; i++){
            squares.push_back(i*i);
        }
        
        bfs.push({0,0});
        while(!bfs.empty()){
            pair<int,int> current_node = bfs.front();
            if(current_node.first==n){
                return current_node.second;
            }
            bfs.pop();
            if(vist[current_node.first]){
               continue; 
            }
            vist[current_node.first]=true;
            for(int i=0; i<100; i++) {
                int suma =current_node.first+squares[i];
                if(suma>n) break;
                bfs.push({suma, current_node.second+1});

            }
        }
        return 0;
    }

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