<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 minNumber(int remain, int val, int nWays, vector<int>& dp)
    {
        if (remain < 0 | val == 0) {
            return INT_MAX;
        }

        if (remain == 0) {
            return nWays;
        }

        if (dp[remain] != INT_MAX) {
            return dp[remain];
        }

        return dp[remain] = min(minNumber(remain - val * val, val, nWays + 1, dp),
                                minNumber(remain, val - 1, nWays, dp));

    }

    int numSquares(int n) {
        int num = sqrt(n);

        if (num * num == n) {
            return 1;
        }
        
        int minValue = INT_MAX;

        for (int i = 1; i <= num; i++) {
            vector<int> dp(n + 1, INT_MAX);
            dp[0] = 0;
            minValue = min(minValue, minNumber(n, i, 0, dp));
        }

        return minValue;
    }
};
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