<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 n;
    int f(int i, int k, vector<int>& arr, vector<vector<int>>&dp){
        if(i>=n) return 0;
        int mx = INT_MIN, ans = INT_MIN, len=0;
        if(dp[i][k]!=-1) return dp[i][k];
        for(int j=i;j<min(i+k,n);j++){
            len++;
            mx = max(arr[j],mx);
            int sum = mx*len + f(j+1,k,arr,dp);
            ans = max(ans, sum);
        }
        return dp[i][k]=ans;
    }

    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        n = arr.size();
        if(n<=k){
            int mx = *max_element(arr.begin(), arr.end());
            return mx*n;
        }
        vector<vector<int>>dp(n,vector<int>(k+1,-1));
        return f(0,k,arr,dp);
    }
};
19000