<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>

#include <cmath>
class Solution {
public:
    int numSquares(int n) {
        


        map<int, bool> m;
        vector<int> vec(n+1, INT_MAX);

        for (int i = 1; i <= n; i++) {
            int root = static_cast<int>(sqrt(i)); 
            if (root * root == i) {
                m.insert({i,true});
            } else {
                m.insert({i,false});
            }
        }

        vec[1] = 1;
        vec[0] = 0;

        for (int i = 2; i  <= n; i++) {

            int local_min = INT_MAX;
            for (int j = 1; j*j <= i; j++) {
                int k = INT_MAX;
                if (m[j*j] == true) {
                    k = 1 + vec[i-j*j];
                   
                }

                if (k < local_min) {
                    local_min = k;
                }

            }
            vec[i] = local_min;
        }

        return vec[n];
    }
};
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