<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:
    def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
        dp = [0]*(1+len(arr))  # dp[r] = max_sum with r elements
        
        for r in range(1, k+1):  # first k elements can be grouped into one subarray
            dp[r] = r * max(arr[:r])

        # for further elements, len(subarray) = 1 to k
        for r in range(k+1, len(dp)):   
            curmax = -1    
            for l in range(r-1, r-k-1, -1):             
                curmax = max(curmax, arr[l])
                dp[r] = max(dp[r], dp[l] + (r-l)*curmax)  # dp[l] + partition = (l+1 to r)
            print(dp)
        return dp[-1]   

17000