<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 func(int ind,int target,vector<int>& arr,vector<vector<int>>& dp){
        if(ind==0){
            if(target%arr[0]==0) return target/arr[0];
            else return 1e9;
        }
        if(dp[ind][target]!=-1) return dp[ind][target];
        int notTake=func(ind-1,target,arr,dp);
        int take=INT_MAX;
        if(target>=arr[ind]){
            take=1+func(ind,target-arr[ind],arr,dp);
        }
        return dp[ind][target]=min(take,notTake);
    }
    int numSquares(int n) {
        vector<int> arr;
        for(int i=1;i*i<=n;i++){
            arr.push_back(i*i);
        }
        int target=n;
        vector<vector<int>>dp(arr.size(),vector<int>(target+1,-1));
        int ans=func(arr.size()-1,target,arr,dp);
        return ans;
    }
};
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