Partition Array for Maximum Sum
<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>
Array; Dynamic Programming
class Solution {
public:
    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        int n = arr.size();
        int f[n + 1];
        memset(f, 0, sizeof(f));
        for (int i = 1; i <= n; ++i) {
            int mx = 0;
            for (int j = i; j > max(0, i - k); --j) {
                mx = max(mx, arr[j - 1]);
                f[i] = max(f[i], f[j - 1] + mx * (i - j + 1));
            }
        }
        return f[n];
    }
};