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0813 - Largest Sum of Averages (Medium)

https://leetcode.com/problems/largest-sum-of-averages/

Problem Statement​

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

Note that the partition must use every integer in nums, and that the score is not necessarily an integer.

Return the maximum score you can achieve of all the possible partitions. Answers within 10-6 of the actual answer will be accepted.

Example 1:

Input: nums = [9,1,2,3,9], k = 3
Output: 20.00000
Explanation: 
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.

Example 2:

Input: nums = [1,2,3,4,5,6,7], k = 4
Output: 20.50000

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i] <= 10^4
  • 1 <= k <= nums.length

Approach 1: Dynamic Programming​

Written by @wingkwong
class Solution {
public:
    double dfs(vector<vector<double>>& dp, vector<int>& nums, int k, int i) {
        // reached all elements
        if (i >= nums.size()) return 0;
        // cannot further partition
        if (k == 0) return INT_MIN;
        // the result has been calculated before
        if (dp[i][k] != -1) return dp[i][k];
        double mx = INT_MIN, sum = 0;
        for (int j = i; j < nums.size(); j++) {
            // sum from nums[i .. j]
            sum += nums[j];
            // current partition + the best result starting from the next index
            mx = max(mx, sum / (j - i + 1) + dfs(dp, nums, k - 1, j + 1));
        }
        // memorise the result
        return dp[i][k] = mx;
    }
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        // The best score partitioning A[i:] into at most K parts
        vector<vector<double>> dp(n, vector<double>(k + 1, -1));
        // starting from index 0
        return dfs(dp, nums, k, 0);
    }
};