813. Largest Sum of Averages

Problem Description

The problem deals with an integer array nums and an integer k. We're tasked with partitioning the array into at most k non-empty adjacent subarrays. Our goal is to maximize the 'score' of the partition, where the score is defined as the sum of the averages of each subarray. Important to note is that we must use every integer in nums and that the partitions we create will affect the score. We need to find the best possible way to split the array to achieve the maximum score.


The core of this problem lies in dynamic programming, particularly in finding the best partition at each stage to maximize the score. A naïve approach might try every possible partition, but this would be prohibitively slow. Instead, we need an efficient way to solve smaller subproblems and combine their solutions to solve the larger problem.

The intuition behind the dynamic programming solution is that if we know the best score we can get from the first i elements with k subarrays, we can use this to compute the best score for the first i+1 elements with k subarrays, and so on. The recursive function dfs represents this notion, where dfs(i, k) returns the best score we can get starting from index i with k subarrays left to form.

The solution uses a top-down approach with memoization (caching) to avoid re-computing the scores for the same i and k values more than once. It also uses a prefix sum array s to quickly compute the sum of elements for any given subarray, which is used to calculate the averages needed to compute the scores for the partitions.

By caching results and avoiding repetition of work, we arrive at the best solution in a much faster and more efficient way than brute-force methods.

Learn more about Dynamic Programming and Prefix Sum patterns.

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1void search(Node root) {
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6    if (!node.visited) {
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9  }

Solution Approach

The solution uses several algorithms and concepts:

Dynamic Programming:

Dynamic programming is utilized to break the problem down into smaller, manageable subproblems. A function dfs recursively computes the maximum score obtainable from a given starting index i and a certain number of partitions k. The essence of dynamic programming is apparent as dfs computes the scores based on previously solved smaller problems.


To optimize the dynamic programming solution, memoization is used. The @cache decorator in Python automatically memorizes the result of the dfs function calls with particular arguments, so when the function is called again with the same arguments, the result is retrieved from the cache instead of re-computing it.

Prefix Sums:

To efficiently calculate the average of any subarray within nums, a prefix sum array s is created using the accumulate function from Python's itertools module, with initial=0 to include the starting point. This data structure allows constant-time retrieval of the sum of elements between any two indices.

Implementation Details:

The main function largestSumOfAverages first computes the cumulative sums of nums. Then it defines the dfs function with parameters i (the starting index for considering partitions) and k (the number of remaining partitions).

The dfs function works as follows:

  • If i == n, where n is the length of nums, it means we've considered all elements and there is no score to be added, so it returns 0.
  • If k == 1, we can only make one more partition, so the best score is the average of all remaining elements, calculated by (s[-1] - s[i]) / (n - i).
  • For other cases, the function iterates through the elements starting from i up to the last element, creating a subarray from i to j and calculating its average. It then recursively calls dfs(j + 1, k - 1) to compute the score of the remaining array with one less partition.
  • The max function keeps track of the highest score found during the iteration.

At the end of the dfs calls, dfs(0, k) provides the maximum score for the entire array with k partitions.

The implementation takes advantage of Python's concise syntax and powerful standard library functions like itertools.accumulate and functools.cache to create an elegant and efficient solution.

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Example Walkthrough

Let's assume we have nums = [9, 1, 2, 3, 9] and k = 3. We want to find the maximum score by partitioning the array into at most k adjacent subarrays.

First, let's calculate the prefix sums to efficiently compute the sums of subarrays:

nums = [9, 1, 2, 3, 9] prefix_sums = [0, 9, 10, 12, 15, 24] (include a 0 at the beginning for easy calculation)

Here's the Python function dfs(i, k) that will compute the maximum score starting from index i with k partitions left.

Now, let's walk through the solution:

Step 1: Call dfs(0, 3) for the full array with 3 partitions allowed.

Step 2: dfs(0, 3) explores partitioning the array from index 0. It tries partitioning after every index to find the maximum score:

  • Partition after index 0: average of first part [9] is 9, now call dfs(1, 2).
  • Partition after index 1: average of first part [9, 1] is 5, now call dfs(2, 2).
  • Partition after index 2: average of first part [9, 1, 2] is 4, now call dfs(3, 2).
  • Partition after index 3: average of first part [9, 1, 2, 3] is 3.75, now call dfs(4, 2).

Step 3: For each of the above calls to dfs(i, 2), it again divides the remaining part of the array and calls dfs(j, 1). When k == 1, it simply takes the average of the remaining elements since it has to be one partition.

Step 4: Each time the score is computed, we take the average of the current partition plus the result of dfs for the remaining partitions. The maximum value from these recursive calls is the answer for the current dfs.

Step 5: Following the steps recursively will lead us to find the maximum score. For instance, one of the optimal solutions is to partition the array into [9], [1, 2, 3], [9] with scores 9 + 2 + 9 = 20.

Step 6: Once all possibilities for dfs(0, 3) are evaluated, the maximum score that can be returned is cached to ensure that the same state is not recomputed.

Step 7: Finally, dfs(0, k) returns the maximum score for the entire array with k partitions.

The recursive and memoizing nature of the dfs function allows us to efficiently explore all possibilities, while the prefix sums provide a quick way to calculate averages as needed without repeated summation. By combining these techniques, the algorithm efficiently finds the solution.

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Python Solution

1from functools import lru_cache
2from itertools import accumulate
3from typing import List
5class Solution:
6    def largestSumOfAverages(self, nums: List[int], k: int) -> float:
7        # Total number of elements in nums
8        num_elements = len(nums)
10        # Prefix sum array including an initial 0 for convenience
11        prefix_sums = list(accumulate(nums, initial=0))
13        # Memoization function for our depth-first search
14        @lru_cache(maxsize=None)
15        def dfs(index, remaining_groups):
16            # Base case: when we have considered all elements
17            if index == num_elements:
18                return 0
19            # When only one group is left, return the average of the remaining elements
20            if remaining_groups == 1:
21                return (prefix_sums[-1] - prefix_sums[index]) / (num_elements - index)
23            max_average_sum = 0
24            # Try to form a group ending at each element from the current index
25            for j in range(index, num_elements):
26                # Current group average sum
27                current_average = (prefix_sums[j + 1] - prefix_sums[index]) / (j - index + 1)
28                # Recursively calculate the sum of averages for the remaining groups
29                total_average_sum = current_average + dfs(j + 1, remaining_groups - 1)
30                # Update the maximum average sum encountered
31                max_average_sum = max(max_average_sum, total_average_sum)
33            return max_average_sum
35        # Launch depth-first search with initial position and group count
36        return dfs(0, k)

Java Solution

1class Solution {
2    private Double[][] memoization; // 2D array for memoization to store intermediate results
3    private int[] prefixSums; // 1D array to store the prefix sums of the input array
4    private int length; // Length of the input array
6    // Calculates the largest sum of averages
7    public double largestSumOfAverages(int[] nums, int k) {
8        length = nums.length;
9        prefixSums = new int[length + 1]; // Array size is length+1 because we start from 1 for easy calculations
10        memoization = new Double[length + 1][k + 1]; // using Double wrapper class to store null initially
11        for (int i = 0; i < length; ++i) {
12            prefixSums[i + 1] = prefixSums[i] + nums[i]; // Compute prefix sums
13        }
14        return dfs(0, k); // Begin depth-first search
15    }
17    // Performs depth-first search to find the maximum sum of averages
18    private double dfs(int startIndex, int groups) {
19        if (startIndex == length) {
20            return 0; // Base case: when we've considered all elements
21        }
22        if (groups == 1) {
23            // If only one group left, return the average of the remaining elements
24            return (double)(prefixSums[length] - prefixSums[startIndex]) / (length - startIndex);
25        }
26        if (memoization[startIndex][groups] != null) {
27            return memoization[startIndex][groups]; // Return cached result if available
28        }
29        double maxAverage = 0;
30        for (int i = startIndex; i < length; ++i) {
31            // Choose different points to split the array
32            double current = (double)(prefixSums[i + 1] - prefixSums[startIndex]) / (i - startIndex + 1) + dfs(i + 1, groups - 1);
33            maxAverage = Math.max(maxAverage, current); // Keep the maximum average found
34        }
35        memoization[startIndex][groups] = maxAverage; // Store the result in memoization array
36        return maxAverage;
37    }

C++ Solution

1#include <vector>
2#include <cstring>
3#include <functional>
5using namespace std;
7class Solution {
9    double largestSumOfAverages(vector<int>& nums, int k) {
10        int n = nums.size();
11        vector<int> prefix_sum(n + 1, 0); // Create a vector to store the prefix sums
12        vector<vector<double>> memo(n, vector<double>(k + 1, 0)); // Create a 2D vector for memoization
14        // Calculate the prefix sums
15        for (int i = 0; i < n; ++i) {
16            prefix_sum[i + 1] = prefix_sum[i] + nums[i];
17        }
19        // Recursive lambda function for depth-first search
20        function<double(int, int)> dfs = [&](int start, int partitions) -> double {
21            if (start == n) return 0; // Base case: no more elements to partition
22            if (partitions == 1) // Base case: only one partition is left
23                return static_cast<double>(prefix_sum[n] - prefix_sum[start]) / (n - start);
25            if (memo[start][partitions] > 0) return memo[start][partitions]; // If value already computed return it
27            double max_average = 0;
28            for (int end = start; end < n; ++end) {
29                double current_average = static_cast<double>(prefix_sum[end + 1] - prefix_sum[start]) / (end - start + 1);
30                double rest_average = dfs(end + 1, partitions - 1);
31                max_average = max(max_average, current_average + rest_average); // Update max_average if the sum of averages is larger
32            }
34            return memo[start][partitions] = max_average; // Memoize and return the result
35        };
37        return dfs(0, k); // Initiate the recursive search with the entire array and k partitions
38    }

Typescript Solution

1// Import required packages for utility functions if needed (TypeScript doesn't use imports in a similar way, but you may need external utility types if you're working outside the Node.js or Deno environment)
3// Define the type alias for 2D array of numbers
4type NumberMatrix = number[][];
6// Define the function to calculate the largest sum of averages
7function largestSumOfAverages(nums: number[], k: number): number {
8    const n: number = nums.length;
9    const prefixSum: number[] = new Array(n + 1).fill(0); // Create an array to store the prefix sums
10    const memo: NumberMatrix = Array.from({length: n}, () => new Array(k + 1).fill(0)); // Create a 2D array for memoization
12    // Calculate the prefix sums
13    for (let i = 0; i < n; ++i) {
14       prefixSum[i + 1] = prefixSum[i] + nums[i];
15    }
17    // Recursive function for depth-first search
18    const dfs: (start: number, partitions: number) => number = (start, partitions) => {
19        if (start === n) return 0; // Base case: no more elements to partition
20        if (partitions === 1) // Base case: only one partition is left
21            return (prefixSum[n] - prefixSum[start]) / (n - start);
23        if (memo[start][partitions] > 0) return memo[start][partitions]; // If value already computed, return it
25        let maxAverage: number = 0;
26        for (let end = start; end < n; ++end) {
27            const currentAverage: number = (prefixSum[end + 1] - prefixSum[start]) / (end - start + 1);
28            const restAverage: number = dfs(end + 1, partitions - 1);
29            maxAverage = Math.max(maxAverage, currentAverage + restAverage); // Update maxAverage if the sum of averages is larger
30        }
32        return memo[start][partitions] = maxAverage; // Memoize and return the result
33    };
35    return dfs(0, k); // Initiate the recursive search with the entire array and k partitions
38// Example usage:
39let nums = [9, 1, 2, 3, 9];
40let k = 3;
41console.log(largestSumOfAverages(nums, k)); // Output will be the result of the function
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Time and Space Complexity

Time Complexity

The time complexity of the given recursive algorithm involves analyzing the number of subproblems solved and the time it takes to solve each subproblem. Here, n represents the length of the nums array, and k represents the number of partitions.

  • There are at most n * k subproblems because for each starting index i in nums, there are at most k partitions possible.
  • For each subproblem defined by a starting index i and a remaining partition count k, the algorithm iterates from i to n-1 to consider all possible partition points.

The time taken for each subproblem is O(n) because of the for-loop from i to n-1. Therefore, the overall time complexity is O(n^2 * k).

Space Complexity

The space complexity consists of the space required by the recursion stack and the caching of subproblem results.

  • The maximum depth of the recursion stack is k, because the algorithm makes a recursive call with k-1 whenever it makes a partition.
  • The cache stores results for the n * k subproblems.

Therefore, the space complexity is O(n * k) due to caching, plus O(k) for the recursion stack, which simplifies to O(n * k).

Learn more about how to find time and space complexity quickly.

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