<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 maxSumAfterPartitioning(vector<int>& arr, int k) {
        int n=arr.size();
        int dp[n][n];
        for(int i=0;i<n;i++){
            int maxi=-1e9;
            for(int j=i;j<n;j++){
                maxi=max(maxi,arr[j]);
                dp[i][j]=maxi*(j-i+1);
            }
        }
        int dp2[n];
        memset(dp2,0,sizeof(dp2));
        for(int i=0;i<n;i++){
            int lim=max(0,i-k+1);
            for(int j=lim;j<=i;j++){
              if(j==0){
                dp2[i]=dp[j][i];
              }
              else{
                dp2[i]=max(dp2[i],dp2[j-1]+dp[j][i]);
              }
            }
        }
        return dp2[n-1];
    }
};
12000