<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:
    vector<int> allSquares;
    vector<int> dp;
    int sz = 0;

    Solution() {
        int i = 1;
        // while(i*i <= 1e4) {
        //     allSquares.push_back(i*i);
        //     i++;
        // }
        // sz = allSquares.size();
        dp.resize(1e4+1, -1);
    }

    int numSquares(int n) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        return solve(n);
    }

    int solve(int n) {
        if(n < 0) {
            return INT_MAX;
        }

        if(n == 0) {
            return 0;
        }

        if(dp[n] != -1) {
            return dp[n];
        }

        int ans = INT_MAX;
        for(int i = 1; i*i <= n; i++) {
            int temp = solve(n - i*i);
            if(temp != INT_MAX) {
                ans = min(ans, temp + 1);
            }
        }

        return dp[n] = ans;
    }
};

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