<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 solve(int index, int n, vector<int>&arr, int k, vector<int>&dp){
    if(index>=n) return 0;
    if(dp[index]!=-1) return dp[index];
    int mx = 0, ans=0;
    for(int i=index;i<index+k && i<n;i++){
        mx = max(mx,arr[i]);
        int p = (mx*(i-index+1))+solve(i+1,n,arr,k,dp);
        ans = max(ans,p);
    }
    return dp[index] = ans;
}
    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        int n = arr.size();
        vector<int>dp(n,-1);
        return solve(0,n,arr,k,dp);
    }
};
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