<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:
    unordered_map<int, int> hmap;
    // for a given state of a number what all reductions that you can do on him and pass it forward
    int solve(int n)
    {
        if(n == 0)return 0;

        if(hmap.find(n) != hmap.end())return hmap[n];

        int result = INT_MAX;

        for(int i = 1; i * i <= n; i++) // the check is also happening here
        {
            // we will try with all the perfect squars less than or equal to n
            int r = 1 + solve(n - (i * i));
            result = min(result, r);
        }

        return hmap[n] = result;
    }

    int numSquares(int n) {
        //return solve(n);

        vector<int> dp(n + 1, 0);

        for(int idx = 1; idx <= n; idx++)
        {
            int result = INT_MAX;

            for(int i = 1; i * i <= idx; i++)
            {
                // we will try with all the perfect squars less than or equal to n
                int r = 1 + solve(idx - (i * i));
                result = min(result, r);
            }

            hmap[idx] = result;
        }

        return hmap[n];
    }
};
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