<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 {
    double helper(vector<vector<double>> &dp,vector<double>& presum,  int index, int k, int n){
        if(index >= n)return 0.0;
        if(k == 0 ) return INT_MIN;
        if(dp[index][k] != 0) return dp[index][k];
        double ans  = 0.0;
        for(int i = index; i < n; i++){ 
            double temp = helper(dp, presum, i + 1, k-1 , n) + (presum[i] - presum[index-1]) / (double)(i - index + 1);
            ans = max(ans, temp);
        }
        dp[index][k] = ans;
        return ans;
        
    }
public:
    double largestSumOfAverages(vector<int>& A, int K) {
        vector<vector<double>> dp(105, vector<double>( 105, 0));
        vector<double> presum;
        presum.push_back(0);
        for(int i = 1; i <= A.size(); i++){
            presum.push_back(presum[i-1] + A[i-1]);
        }
        return helper(dp,presum,1,K,presum.size());
    }
};
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