<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>v;
    bool isPerfectSquare(int n){
        int s = sqrt(n);
        return (n==(s*s));
    }
    int solve(int i,int n,vector<vector<int>>&dp){
        if(i>=v.size() or n<0){
            return 1e9;
        }
        if(n==v[i]){
            return 1;
        }
        if(dp[i][n] != -1){
            return dp[i][n];
        }
        int notTake = solve(i+1,n,dp);
        int take=1e9;
        if(v[i]<=take){
            take = 1 + solve(i,n-v[i],dp);
        }

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


    int numSquares(int n) {
        for(int i=1;i<=n;i++){
            if(isPerfectSquare(i)){
                v.push_back(i);
            }
        }
        if(isPerfectSquare(n)){
            return 1;
        }
        vector<vector<int>>dp(v.size()+1,vector<int>(n+1,-1));
        sort(v.begin(),v.end(),greater<int>());
        int ans = solve(0,n,dp);
        return ans;
    }
};
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