<p>Given an integer array <code>arr</code>, partition the array into (contiguous) subarrays of length <strong>at most</strong> <code>k</code>. After partitioning, each subarray has their values changed to become the maximum value of that subarray.</p>

<p>Return <em>the largest sum of the given array after partitioning. Test cases are generated so that the answer fits in a <strong>32-bit</strong> integer.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,15,7,9,2,5,10], k = 3
<strong>Output:</strong> 84
<strong>Explanation:</strong> arr becomes [15,15,15,9,10,10,10]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
<strong>Output:</strong> 83
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [1], k = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 500</code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= arr.length</code></li>
</ul>

class Solution {
public:


    int recursion(vector<int> &arr , int i, int k ,vector<int>&dp){
        
        if(i>=arr.size())return 0;
        if(dp[i]!=-1)return dp[i];
        int currmax =INT_MIN; 
        int ans =INT_MIN; 
        for(int j = 0 ;j<k;j++){
            if(i+j>=arr.size())break;
            currmax=max(currmax,arr[i+j]);
            ans = max(ans , currmax*(j+1)+recursion(arr,i+j+1,k,dp));
        }
        dp[i]=ans;
        return ans;
    }

    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        
        vector<int> dp(arr.size(),-1);
        return recursion(arr,0,k,dp);
    }
};
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