<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. You can partition the array into <strong>at most</strong> <code>k</code> non-empty adjacent subarrays. The <strong>score</strong> of a partition is the sum of the averages of each subarray.</p>

<p>Note that the partition must use every integer in <code>nums</code>, and that the score is not necessarily an integer.</p>

<p>Return <em>the maximum <strong>score</strong> you can achieve of all the possible partitions</em>. Answers within <code>10<sup>-6</sup></code> of the actual answer will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [9,1,2,3,9], k = 3
<strong>Output:</strong> 20.00000
<strong>Explanation:</strong> 
The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 4
<strong>Output:</strong> 20.50000
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution {
public:
    double largestSumOfAverages(vector<int>& nums, int k) 
    {
      // It is classic DP problem! so we will define dp[i][k] as best score of all elements till index i with exactly k partitions!
      int n = nums.size();
      vector<vector<double>> dp(n+5,vector<double>(n+5,0.0));
      double sum = 0;
      for(int i=1;i<=n;i++)
      {
        sum += nums[i-1];
        dp[i][1] = (double)sum/i;
      }
      for(int i=2;i<=n;i++)
      {
        for(int part=2;part<=k;part++)
        {
            int j = i;
            double sum = 0;
            while(j>=1)
            {
              int number_of_elements = i - j + 1;
              sum = sum +  (double)nums[j-1];
              double opt = (double)sum/number_of_elements + dp[j-1][part-1];
              dp[i][part] = max(dp[i][part],opt);
              j--;
            }
        }
      }
      return dp[n][k];
    }
};
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