<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 helper(vector<int> &nums, int k, int idx, vector<vector<double>> &dp) {
        if (k < 0 || idx >= nums.size()) return 0;
        if(k==0)return INT_MIN/2;
        
        if (dp[k][idx] != -1) return dp[k][idx];
        
        // If no more partitions are allowed, return the average of the remaining segment
        if (k == 0) {
            return (idx > 0) ? (nums.back() - nums[idx - 1]) / (nums.size() - idx) : (double) nums.back() / nums.size();
        }
        
        double maxAvg = 0;
        for (int i = idx; i < nums.size(); i++) {
            double segmentSum = (idx > 0) ? nums[i] - nums[idx - 1] : nums[i]; // Sum of elements from idx to i
            double avg = segmentSum / (i - idx + 1);
            maxAvg = max(maxAvg, avg + helper(nums, k - 1, i + 1, dp));
        }
        
        return dp[k][idx] = maxAvg;
    }

    double largestSumOfAverages(vector<int>& nums, int k) {
        // Create prefix sum array
        for (int i = 1; i < nums.size(); i++) {
            nums[i] += nums[i - 1];
        }

        // dp[k][idx]: maximum sum of averages we can get with k partitions starting from idx
        vector<vector<double>> dp(k + 1, vector<double>(nums.size(), -1));
        
        // Start the helper function
        return helper(nums, k, 0, dp);
    }
};

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