<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> vec(n+1,INT_MAX);
        // vec[0] = 0;
        // int count = 1;
        // while( count*count <= n ) {
        //     int sq = count*count;
        //     for(int i = sq; i < n+1; i++) {
        //         vec[i] = min(vec[i-sq]+1, vec[i]);
        //     }
        //     count++;
        // }

        // return vec[n];

        vector<int> squares;
        for(int i = 1; i*i<=n; i++)
            squares.push_back(i*i);
        int step = 0;

        queue<int> q;
        q.push(n);
        while(!q.empty()) {
            step++;
            queue<int> temp_q;
            for(int size = q.size(); size > 0; size--) {
                int num = q.front();
                q.pop();
                for(auto sq : squares) {
                    if (sq == num)
                        return step;
                    if (sq > num)
                        break;
                    temp_q.push(num-sq);
                }
            }
            q = temp_q;
            // step++;
        } // while

        return step;
    }
};
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