<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 fun(int i, vector<int>& nums, int n, vector<vector<int>>& dp) {
    if (n == 0) return 0;
    if (n < 0) return 1e8;
    if (i < 0) return 1e8;
    if (dp[i][n] != -1) return dp[i][n];

    int take = 1e8;
    if (nums[i] <= n) take = 1 + fun(i, nums, n - nums[i], dp);
    int not_take = fun(i - 1, nums, n, dp);

    return dp[i][n] = min(take, not_take);
}

int numSquares(int n) {
    vector<int> nums;
    for (int i = 1; i * i <= n; ++i) {
        nums.push_back(i * i);
    }
    int m = nums.size();
    vector<vector<int>> dp(m, vector<int>(n + 1, -1));

    return fun(m - 1, nums, n, dp);
}
};

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