Introduction

In this blog post, we will explore how to solve the LeetCode problem "813" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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] <= 104 1 <= k <= nums.length

Explanation

Here's the solution to the problem:

• Dynamic Programming: Use dynamic programming to store and reuse the maximum scores achievable for different subarray partitions.
• Prefix Sums: Utilize prefix sums to efficiently calculate the average of subarrays in O(1) time.

• Iterative Approach: Build the solution iteratively, considering all possible partition points for each number of subarrays (from 1 to k).

• Runtime Complexity: O(n*n*k), where n is the length of nums and k is the maximum number of subarrays. Storage Complexity: O(n*k).

Code

    def largestSumOfAverages(nums, k):
    n = len(nums)
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + nums[i]

    dp = [[0.0] * (k + 1) for _ in range(n + 1)]

    for i in range(1, n + 1):
        dp[i][1] = prefix_sum[i] / i

    for j in range(2, k + 1):
        for i in range(j, n + 1):
            for x in range(j - 1, i):
                dp[i][j] = max(dp[i][j], dp[x][j - 1] + (prefix_sum[i] - prefix_sum[x]) / (i - x))

    return dp[n][k]