<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 helper(int root, int target, vector<vector<int>>& dp){
        if(target == 0) return 0;
        if(root < 1) return 1e5;

        if(dp[root][target] != -1) return dp[root][target];

        int take = INT_MAX, not_take = INT_MAX;
        
        if((root * root) <= target) take = 1 + helper(root, target - (root * root), dp);
        not_take = helper(root - 1, target, dp);

        return dp[root][target] = min(take, not_take);
    }

    int numSquares(int n) {
        int start = 1, end = n;

        if(n == 1) return 1;

        while(start < end){
            int mid = (start + end)/2;

            if((mid * mid) == n){
                return 1;
            }
            else if((mid * mid) < n) start += 1;
            else end = mid;
        }

        vector<vector<int>> dp(start, vector<int> (n + 1, -1));
        return helper(start - 1, n, dp);
    }
};
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