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 <= 1001 <= nums[i] <= 1041 <= k <= nums.lengthThe key idea in #813 Largest Sum of Averages is to split the array into at most k contiguous groups such that the sum of their averages is maximized. A brute force approach would try every possible partition, but that becomes computationally expensive as the number of combinations grows quickly.
An efficient strategy uses prefix sums and dynamic programming. Prefix sums allow you to compute the average of any subarray in constant time using (prefix[j] - prefix[i]) / (j - i). Then define a DP state where dp[i][g] represents the maximum sum of averages achievable using the first i elements divided into g groups.
For each state, iterate over possible previous partition points and update the best result by combining earlier DP values with the average of the current segment. This structured exploration significantly reduces redundant work. The approach typically runs in O(n^2 * k) time with O(n * k) space.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Dynamic Programming with Prefix Sum | O(n^2 * k) | O(n * k) |
| Prefix Sum Preprocessing | O(n) | O(n) |
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This approach utilizes dynamic programming combined with prefix sums to efficiently calculate the sum of subarray averages when partitioned optimally. The prefix sum array allows for quick computation of subarray sums, and dynamic programming is used to build the optimal solution up to k partitions.
Time Complexity: O(n^2 * k), where n is the length of the array, and k is the number of partitions.
Space Complexity: O(n * k) due to the dp table.
1#include <stdio.h>
2#include <string.h>
3#include <math.h>
4
5#define MAX_N 100
6
First, compute prefix sums for the array, then utilize dynamic programming to calculate the maximum sum of averages for each possible partition using the prefix sums. The dp array stores the maximum score achievable for each subarray starting at index i using up to m partitions.
This approach uses recursion with memoization to solve for the maximum sum of averages by exploring all possible partition strategies and storing computed results for reuse. This avoids redundant computations and optimizes performance compared to plain recursion.
Time Complexity: O(n^2 * k) as each subproblem is solved once.
Space Complexity: O(n * k) for memoization table.
1#
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Prefix sums help compute the sum of any subarray quickly, which makes calculating averages efficient. Without them, repeatedly summing subarrays would significantly increase the time complexity.
Yes, problems like Largest Sum of Averages are commonly used in technical interviews at large tech companies. They test understanding of dynamic programming, partitioning strategies, and optimization using prefix sums.
The optimal approach uses dynamic programming combined with prefix sums. Prefix sums allow constant-time subarray average calculations, while DP tracks the best score when dividing the array into different numbers of groups.
A dynamic programming table is the main data structure used. It stores the best possible sum of averages for different prefixes of the array and group counts, while a prefix sum array helps compute segment averages efficiently.
This C implementation defines a solve function that handles partitioning recursively, memoizing previously calculated scores to prevent redundant computation. This function checks scores for partitions from index i with m partitions left.