<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 bfs(vector<int>p,int t){
        queue<pair<int,int>>q;
        q.push(make_pair(0,0));
        unordered_set<int>s;
        while(!q.empty()){
            pair<int,int> k=q.front();
            q.pop();
            for(auto i:p){
                int sum=k.first+i;
                if(sum==t){
                    return k.second+1;
                }
                if(sum<t && !s.count(sum)){
                    q.push(make_pair(sum,k.second+1));
                } 
            }
        }
        return 1;
    }
    int numSquares(int n) {
        vector<int>g;
        for(int i=1;i<=n;i++){
            int h=sqrt(i);
            if(h*h==i){
                g.push_back(i);
            }
        }
        int c=bfs(g,n);
        return c;

    }
};
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