<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) {
        // int ans = 0;
        vector<int> dp(n+1,INT_MAX);
        vector<int> squares;
        for(int i = 1; i*i<= n; i++){
            squares.push_back(i*i);
        }
        int m = squares.size();
        int ans = 1;
        queue<int> q;
        q.push(n);
        while(q.size()){
            int k = q.size();
            for(int i = 0;i<k;i++){
                auto u = q.front();
                q.pop();
                for(int j = 0;j < m;j++){
                    if(squares[j] > u)break;
                    if(squares[j] == u)return ans;
                    q.push(u - squares[j]);
                }
            }
            ans++;
        }
        return ans;
        // vector<int> dp(n+1,INT_MAX);
        // dp[1] = 1;
        // for(int i = 2;i<=n;i++){
        //     int ub = sqrt(i);
        //     if(ub*ub == i){
        //         dp[i] = 1;
        //         continue;
        //     }
        //     for(int j = 1;j<=(i/2);j++){
        //         dp[i] = min(dp[i], dp[j] + dp[i-j]);
        //     }
        // }
        // // for(auto i : dp)cout << i << " ";
        // return dp[n];
    }
};
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