<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 rec(self, ind, arr, n, k, dp):
        if ind == n: return 0
        if dp[ind] != -1: return dp[ind]
        maxi = -1e9
        maxSum = -1e9
        length = 0
        for j in range(ind, min(ind + k, n)):
            length += 1
            maxi = max(maxi, arr[j])
            sm = length * maxi + self.rec(j + 1, arr, n, k, dp)
            maxSum = max(sm, maxSum)
        dp[ind] = maxSum
        return dp[ind]

    def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
        n = len(arr)
        dp = [-1]*n
        return self.rec(0, arr, n, k, dp)
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