problemmediumalgorithmsleetcode-813leetcode 813leetcode813

Largest Sum of Averages

MediumUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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.

Examples

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

Solution

Method 1 - Top Down DFS (Memoized)

Intuition

To maximize the sum of averages when partitioning the array into at most k groups, we want to split the array so that each group has a high average. This means larger numbers should ideally be in their own group, while smaller numbers can be grouped together. Dynamic programming helps us efficiently explore all possible ways to split the array and keep track of the best result for each subproblem.

Approach

We use a recursive DFS to partition the array into at most k groups, maximizing the sum of averages. For each partition, we try all possible splits and use memoization to cache results for subproblems. The function search(n, k) returns the largest sum of averages for the first n elements split into k groups.

Recurrence Relation

Let dp(n, k) be the maximum sum of averages for the first n elements split into k groups.

  • If k == 1, dp(n, 1) = average(nums[0..n-1])
  • Otherwise, dp(n, k) = max(dp(i, k-1) + average(nums[i..n-1])) for all i in [k-1, n-1]
Base Case

If k == 1, the answer is the average of the first n elements.

Code

C++
class Solution {
public:
	double largestSumOfAverages(vector<int>& nums, int k) {
		int n = nums.size();
		vector<vector<double>> dp(n + 1, vector<double>(k + 1, 0.0));
		vector<double> prefix(n + 1, 0.0);
		for (int i = 0; i < n; ++i)
			prefix[i + 1] = prefix[i] + nums[i];
		for (int i = 1; i <= n; ++i)
			dp[i][1] = prefix[i] / i;
		for (int g = 2; g <= k; ++g) {
			for (int i = g; i <= n; ++i) {
				for (int j = g - 1; j < i; ++j) {
					dp[i][g] = max(dp[i][g], dp[j][g - 1] + (prefix[i] - prefix[j]) / (i - j));
				}
			}
		}
		return dp[n][k];
	}
};
Go
func largestSumOfAverages(nums []int, k int) float64 {
	n := len(nums)
	dp := make([][]float64, n+1)
	for i := range dp {
		dp[i] = make([]float64, k+1)
	}
	prefix := make([]float64, n+1)
	for i := 0; i < n; i++ {
		prefix[i+1] = prefix[i] + float64(nums[i])
	}
	for i := 1; i <= n; i++ {
		dp[i][1] = prefix[i] / float64(i)
	}
	for g := 2; g <= k; g++ {
		for i := g; i <= n; i++ {
			for j := g - 1; j < i; j++ {
				avg := (prefix[i] - prefix[j]) / float64(i-j)
				if dp[j][g-1]+avg > dp[i][g] {
					dp[i][g] = dp[j][g-1] + avg
				}
			}
		}
	}
	return dp[n][k]
}
Java
class Solution {
	public double largestSumOfAverages(int[] nums, int k) {
		int n = nums.length;
		double[][] dp = new double[n + 1][k + 1];
		double[] prefix = new double[n + 1];
		for (int i = 0; i < n; ++i)
			prefix[i + 1] = prefix[i] + nums[i];
		for (int i = 1; i <= n; ++i)
			dp[i][1] = prefix[i] / i;
		for (int g = 2; g <= k; ++g) {
			for (int i = g; i <= n; ++i) {
				for (int j = g - 1; j < i; ++j) {
					dp[i][g] = Math.max(dp[i][g], dp[j][g - 1] + (prefix[i] - prefix[j]) / (i - j));
				}
			}
		}
		return dp[n][k];
	}
}
Kotlin
class Solution {
	fun largestSumOfAverages(nums: IntArray, k: Int): Double {
		val n = nums.size
		val dp = Array(n + 1) { DoubleArray(k + 1) }
		val prefix = DoubleArray(n + 1)
		for (i in nums.indices) prefix[i + 1] = prefix[i] + nums[i]
		for (i in 1..n) dp[i][1] = prefix[i] / i
		for (g in 2..k) {
			for (i in g..n) {
				for (j in (g - 1) until i) {
					dp[i][g] = maxOf(dp[i][g], dp[j][g - 1] + (prefix[i] - prefix[j]) / (i - j))
				}
			}
		}
		return dp[n][k]
	}
}
Python
class Solution:
	def largestSumOfAverages(self, nums: List[int], k: int) -> float:
		n = len(nums)
		prefix = [0] * (n + 1)
		for i in range(n):
			prefix[i + 1] = prefix[i] + nums[i]
		dp = [[0.0] * (k + 1) for _ in range(n + 1)]
		for i in range(1, n + 1):
			dp[i][1] = prefix[i] / i
		for g in range(2, k + 1):
			for i in range(g, n + 1):
				for j in range(g - 1, i):
					dp[i][g] = max(dp[i][g], dp[j][g - 1] + (prefix[i] - prefix[j]) / (i - j))
		return dp[n][k]
Rust
impl Solution {
	pub fn largest_sum_of_averages(nums: Vec<i32>, k: i32) -> f64 {
		let n = nums.len();
		let mut prefix = vec![0.0; n + 1];
		for i in 0..n {
			prefix[i + 1] = prefix[i] + nums[i] as f64;
		}
		let k = k as usize;
		let mut dp = vec![vec![0.0; k + 1]; n + 1];
		for i in 1..=n {
			dp[i][1] = prefix[i] / i as f64;
		}
		for g in 2..=k {
			for i in g..=n {
				for j in (g - 1)..i {
					dp[i][g] = dp[i][g].max(dp[j][g - 1] + (prefix[i] - prefix[j]) / (i - j) as f64);
				}
			}
		}
		dp[n][k]
	}
}
TypeScript
class Solution {
	largestSumOfAverages(nums: number[], k: number): number {
		const n = nums.length;
		const prefix: number[] = Array(n + 1).fill(0);
		for (let i = 0; i < n; ++i) prefix[i + 1] = prefix[i] + nums[i];
		const dp: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
		for (let i = 1; i <= n; ++i) dp[i][1] = prefix[i] / i;
		for (let g = 2; g <= k; ++g) {
			for (let i = g; i <= n; ++i) {
				for (let j = g - 1; j < i; ++j) {
					dp[i][g] = Math.max(dp[i][g], dp[j][g - 1] + (prefix[i] - prefix[j]) / (i - j));
				}
			}
		}
		return dp[n][k];
	}
}

Complexity

  • ⏰ Time complexity: O(KN^2), because for each group count g and each end index i, we try all possible previous split points j.
  • 🧺 Space complexity: O(KN), for the DP table storing results for all subproblems.

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