<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> squares;
        for(int i = 1; i <= n; i++)
        {
            if(isPerfectSquare(i)) squares.push_back(i);
        }
        
        queue<pair<int, int>> sums;
        sums.push({0, 0});
        
        while(!sums.empty())
        {
            auto [sum, count] = sums.front();
            sums.pop();
            
            for(const int& square : squares)
            {
                int currSum = sum + square;
                if(currSum <= n)
                {
                    if(currSum == n) return count + 1;
                    sums.push({currSum, count + 1});
                }
            }
        }
        
        return -1;
    }
    
    bool isPerfectSquare(int num) {
         // Calculate square root
        int root = sqrt(num);
        
        // Check if square root squared equals the original number
        return root * root == num; 
    }
};
91283