<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 ind,int tar,vector<long>& arr,vector<vector<int>>&dp){
        int n=arr.size();
        if(tar==0) return 1;
        if(ind==n-1){
            if(tar%arr[ind]==0) return tar/arr[ind];
            else return 1e9;
        }
        if(dp[ind][tar]!=-1) return dp[ind][tar];
        int not_take=solve(ind+1,tar,arr,dp);
        int take=1e9;
        if(tar>=arr[ind])
        take=1+solve(ind,tar-arr[ind],arr,dp);

        return dp[ind][tar]=min(take,not_take);
    }
    int numSquares(int n) {
        if(n==1) return 1;
        vector<long>arr;
        int num=1;
        long sqr=1;
        while(sqr<n){
            sqr=num*num;
            arr.push_back(sqr);
            num++;
        }
        int m=arr.size();
        for(int i=0;i<m;i++) if(arr[i]==n) return 1;

        vector<vector<int>>dp(m,vector<int>(n+1,-1));
        return solve(0,n,arr,dp)-1;
    }
};
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