813. Largest Sum of Averages

Description

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] <= 10⁴
1 <= k <= nums.length

Solutions

Solution 1: Prefix Sum + Memoized Search

We can preprocess to obtain the prefix sum array $s$, which allows us to quickly get the sum of subarrays.

Next, we design a function $\textit{dfs}(i, k)$, which represents the maximum sum of averages when dividing the array starting from index $i$ into at most $k$ groups. The answer is $\textit{dfs}(0, k)$.

The execution logic of the function $\textit{dfs}(i, k)$ is as follows:

When $i = n$, it means we have traversed to the end of the array, and we return $0$.
When $k = 1$, it means there is only one group left, and we return the average value from index $i$ to the end of the array.
Otherwise, we enumerate the starting position $j$ of the next group in the interval $[i + 1, n)$, calculate the average value from $i$ to $j - 1$ as $\frac{s[j] - s[i]}{j - i}$, add the result of $\textit{dfs}(j, k - 1)$, and take the maximum value of all results.

The time complexity is $O(n^2 \times k)$, and the space complexity is $O(n \times k)$. Here, $n$ represents the length of the array $\textit{nums}$.

Python3JavaC++GoTypeScript

class Solution:
    def largestSumOfAverages(self, nums: List[int], k: int) -> float:
        @cache
        def dfs(i: int, k: int) -> float:
            if i == n:
                return 0
            if k == 1:
                return (s[n] - s[i]) / (n - i)
            ans = 0
            for j in range(i + 1, n):
                ans = max(ans, (s[j] - s[i]) / (j - i) + dfs(j, k - 1))
            return ans

        n = len(nums)
        s = list(accumulate(nums, initial=0))
        return dfs(0, k)

class Solution {
    private Double[][] f;
    private int[] s;
    private int n;

    public double largestSumOfAverages(int[] nums, int k) {
        n = nums.length;
        s = new int[n + 1];
        f = new Double[n][k + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        return dfs(0, k);
    }

    private double dfs(int i, int k) {
        if (i == n) {
            return 0;
        }
        if (k == 1) {
            return (s[n] - s[i]) * 1.0 / (n - i);
        }
        if (f[i][k] != null) {
            return f[i][k];
        }
        double ans = 0;
        for (int j = i + 1; j < n; ++j) {
            ans = Math.max(ans, (s[j] - s[i]) * 1.0 /(j - i) + dfs(j, k - 1));
        }
        return f[i][k] = ans;
    }
}

class Solution {
public:
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        int s[n + 1];
        double f[n][k + 1];
        memset(f, 0, sizeof(f));
        s[0] = 0;
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        auto dfs = [&](auto&& dfs, int i, int k) -> double {
            if (i == n) {
                return 0;
            }
            if (k == 1) {
                return (s[n] - s[i]) * 1.0 / (n - i);
            }
            if (f[i][k] > 0) {
                return f[i][k];
            }
            double ans = 0;
            for (int j = i + 1; j < n; ++j) {
                ans = max(ans, (s[j] - s[i]) * 1.0 / (j - i) + dfs(dfs, j, k - 1));
            }
            return f[i][k] = ans;
        };
        return dfs(dfs, 0, k);
    }
};

func largestSumOfAverages(nums []int, k int) float64 {
    n := len(nums)
    s := make([]int, n+1)
    for i, x := range nums {
        s[i+1] = s[i] + x
    }
    f := make([][]float64, n)
    for i := range f {
        f[i] = make([]float64, k+1)
    }
    var dfs func(int, int) float64
    dfs = func(i, k int) float64 {
        if i == n {
            return 0
        }
        if f[i][k] > 0 {
            return f[i][k]
        }
        if k == 1 {
            return float64(s[n]-s[i]) / float64(n-i)
        }
        ans := 0.0
        for j := i + 1; j < n; j++ {
            ans = math.Max(ans, float64(s[j]-s[i])/float64(j-i)+dfs(j, k-1))
        }
        f[i][k] = ans
        return ans
    }
    return dfs(0, k)
}

function largestSumOfAverages(nums: number[], k: number): number {
    const n = nums.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 0; i < n; i++) {
        s[i + 1] = s[i] + nums[i];
    }
    const f: number[][] = Array.from({ length: n }, () => Array(k + 1).fill(0));
    const dfs = (i: number, k: number): number => {
        if (i === n) {
            return 0;
        }
        if (f[i][k] > 0) {
            return f[i][k];
        }
        if (k === 1) {
            return (s[n] - s[i]) / (n - i);
        }
        for (let j = i + 1; j < n; j++) {
            f[i][k] = Math.max(f[i][k], dfs(j, k - 1) + (s[j] - s[i]) / (j - i));
        }
        return f[i][k];
    };
    return dfs(0, k);
}

Solution 2: Dynamic Programming

We can transform the memoized search from Solution 1 into dynamic programming.

Define $f[i][j]$ to represent the maximum sum of averages when dividing the first $i$ elements of the array $\textit{nums}$ into at most $j$ groups. The answer is $f[n][k]$.

For $f[i][j]$, we can enumerate the end position $h$ of the previous group, calculate $f[h][j-1]$, add the result of $\frac{s[i] - s[h]}{i - h}$, and take the maximum value of all results.