<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:
   
    unordered_map<int,int> mp;
    int solve(vector<int>& arr, int k, int ind){
        if(ind>=arr.size()){
            return 0;
        }
        if(mp[ind]!=0){
            return mp[ind];
        }
        int ans=INT_MIN,len=0,maxi= INT_MIN,n=arr.size();
        for(int i=ind; i<min(n,ind+k); i++){
            len=len+1;
            maxi=max(maxi,arr[i]);
            int temp= len*maxi+ solve(arr,k,i+1);
            ans=max(ans,temp);
        }

        return mp[ind]=ans;
    }
public:
    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        return solve(arr,k,0);
    }
};
13000