<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 dp[501][501];
    int solve(int idx,vector<int>&arr, int k){
        int n=arr.size();
        if(idx>=arr.size()){
            return 0;
        }
        if(dp[idx][k]!=-1){
            return dp[idx][k];
        }
        int length=0;
        int sum=0;
         int maxi=INT_MIN;
        int maxAns=INT_MIN;
        for(int p=idx;p<min(n,idx+k);p++){
            length++;
            maxi=max(maxi,arr[p]);
            sum=length*maxi+solve(p+1,arr,k);
            maxAns=max(maxAns,sum);
        }
        return dp[idx][k]= maxAns;
    }
    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        memset(dp,-1,sizeof(dp));
         return solve(0,arr,k);

    }
};
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