<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 {
private:

    int GetSum(const vector<int>& arr, int& i, int& j)
    {
        int maxi = *max_element(arr.begin() + i, arr.begin() + j + 1);
        return maxi * (j - i + 1);
    }
    int f(const vector<int>& arr, const int& k, const int& n, int ind, vector<int>& dp)
    {
        if(ind >= n)
            return 0;

        if(dp[ind] != -1)
            return dp[ind];

        int maxi = INT_MIN;
        for(int x = ind; x < min(n, ind + k); x++)
        {
            int cost = GetSum(arr, ind, x) + f(arr, k, n, x + 1, dp);
            maxi = max(cost, maxi);
        }

        return dp[ind] = maxi;
    }
public:
    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        int n = arr.size();
        vector<int> dp(n, -1);
        return f(arr, k, n, 0, dp);
    }
};
12500