<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 solve(int i, vector<int>& arr, int sum, int target, vector<vector<int>>& dp) {
        if (sum == target) return 0;
        if (i >= arr.size() || sum > target) return 1e9;
        if (dp[i][sum] != -1) return dp[i][sum];
        int take = 1 + solve(i, arr, sum + arr[i], target, dp); //1+ for count
        int not_take = solve(i + 1, arr, sum, target, dp);
        return dp[i][sum] = min(take, not_take);
        //i for which perfect square is taken and the current sum assosiated with it
    }
    
    int numSquares(int n) {
        vector<int> arr; //array of perfect squares
        for (int i = 1; i * i <= n; i++) {
            arr.push_back(i * i);
        }
        vector<vector<int>> dp(arr.size(), vector<int>(n + 1, -1));
        
        return solve(0, arr, 0, n, dp);
    }
};

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