<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 {
private:
int solve(vector<int>& arr, int k,int index,vector<int> &dp){
    if(index==arr.size()){
        return 0;
    }
    if(dp[index] != -1){
        return dp[index];
    }
    int maxi=INT_MIN;
    int maxans=INT_MIN;
    int length=0;
    for(int i=index;i<min((int)arr.size(),index+k);i++){
        maxi=max(maxi,arr[i]);
        length++;
        int sum=(maxi*length)+solve(arr,k,i+1,dp);
        maxans=max(maxans,sum);
    }
    return dp[index]=maxans;
}
public:
    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        vector<int> dp(arr.size()+1,-1);
        return solve(arr,k,0,dp);
    }
};
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