Introduction

In this blog post, we will explore how to solve the LeetCode problem "3473" 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 two integers, k and m. Return the maximum sum of k non-overlapping subarrays of nums, where each subarray has a length of at least m. Example 1: Input: nums = [1,2,-1,3,3,4], k = 2, m = 2 Output: 13 Explanation: The optimal choice is: Subarray nums[3..5] with sum 3 + 3 + 4 = 10 (length is 3 >= m). Subarray nums[0..1] with sum 1 + 2 = 3 (length is 2 >= m). The total sum is 10 + 3 = 13. Example 2: Input: nums = [-10,3,-1,-2], k = 4, m = 1 Output: -10 Explanation: The optimal choice is choosing each element as a subarray. The output is (-10) + 3 + (-1) + (-2) = -10. Constraints: 1 <= nums.length <= 2000 -104 <= nums[i] <= 104 1 <= k <= floor(nums.length / m) 1 <= m <= 3

Explanation

Here's a solution that addresses the problem efficiently:

• Dynamic Programming: Use a 3D DP table dp[i][j][l] to store the maximum sum achievable using j subarrays considering elements up to index i, where l indicates whether the last element is part of a chosen subarray or not.
• Iteration and Transition: Iterate through the array and the number of subarrays. Update the DP table based on whether we extend a previous subarray, start a new one, or skip the current element. The condition i >= (j * m -1) ensures there are sufficient elements to form j subarrays.

• Optimization: Since the last element can be either included or excluded from a subarray, the value of 'l' is just 0 or 1.

• Runtime Complexity: O(n k m), where n is the length of nums, k is the number of subarrays, and m is the minimum length of each subarray. Storage Complexity: O(n k 2) which simplifies to O(n*k).

Code

    def maxSumSubarrays(nums, k, m):
    n = len(nums)

    # dp[i][j][l]: max sum using j subarrays up to index i,
    # where l=1 if nums[i] is part of a subarray, l=0 otherwise
    dp = [[[float('-inf')] * 2 for _ in range(k + 1)] for _ in range(n + 1)]

    # Base case: 0 subarrays with 0 elements have a sum of 0
    dp[0][0][0] = 0

    for i in range(1, n + 1):
        for j in range(k + 1):
            # Option 1: Don't include nums[i-1] in any subarray
            dp[i][j][0] = max(dp[i-1][j][0], dp[i-1][j][1])

            # Option 2: Include nums[i-1] as part of an existing subarray
            if j > 0:
                current_sum = 0
                for length in range(1, i + 1):  # Iterate through possible lengths of the last subarray
                    if length >= m:
                        current_sum = sum(nums[i-length:i])
                        if i-length == 0:
                           dp[i][j][1] = max(dp[i][j][1],current_sum)
                        else:
                            dp[i][j][1] = max(dp[i][j][1], dp[i-length][j-1][0] + current_sum, dp[i-length][j-1][1] + current_sum)


    return max(dp[n][k][0], dp[n][k][1])