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Maximum Score Of Spliced Array

You are given two 0-indexed integer arrays nums1 and nums2, both of length n.

You can choose two integers left and right where 0 <= left <= right < n and swap the subarray nums1[left...right] with the subarray nums2[left...right].

  • For example, if nums1 = [1,2,3,4,5] and nums2 = [11,12,13,14,15] and you choose left = 1 and right = 2, nums1 becomes [1,12,13,4,5] and nums2 becomes [11,2,3,14,15].

You may choose to apply the mentioned operation once or not do anything.

The score of the arrays is the maximum of sum(nums1) and sum(nums2), where sum(arr) is the sum of all the elements in the array arr.

Return the maximum possible score.

A subarray is a contiguous sequence of elements within an array. arr[left...right] denotes the subarray that contains the elements of nums between indices left and right (inclusive).

 

Example 1:

Input: nums1 = [60,60,60], nums2 = [10,90,10]
Output: 210
Explanation: Choosing left = 1 and right = 1, we have nums1 = [60,90,60] and nums2 = [10,60,10].
The score is max(sum(nums1), sum(nums2)) = max(210, 80) = 210.

Example 2:

Input: nums1 = [20,40,20,70,30], nums2 = [50,20,50,40,20]
Output: 220
Explanation: Choosing left = 3, right = 4, we have nums1 = [20,40,20,40,20] and nums2 = [50,20,50,70,30].
The score is max(sum(nums1), sum(nums2)) = max(140, 220) = 220.

Example 3:

Input: nums1 = [7,11,13], nums2 = [1,1,1]
Output: 31
Explanation: We choose not to swap any subarray.
The score is max(sum(nums1), sum(nums2)) = max(31, 3) = 31.

 

Constraints:

  • n == nums1.length == nums2.length
  • 1 <= n <= 105
  • 1 <= nums1[i], nums2[i] <= 104

Solution Explanation

This problem asks to find the maximum score achievable by swapping a subarray between two arrays nums1 and nums2. The score is the maximum of the sums of the modified nums1 and nums2. The optimal solution leverages the concept of maximum subarray sum to efficiently find the best subarray to swap.

Core Idea:

Instead of directly iterating through all possible subarray swaps (which would be O(n^3)), we observe that the effect of swapping a subarray [left, right] is simply adding the difference between the sums of the subarrays from nums2 and nums1 to the original sum of nums1 (or subtracting it from the original sum of nums2). This difference can be calculated efficiently using Kadane's algorithm for finding the maximum subarray sum.

Algorithm:

  1. Calculate initial sums: Calculate the sum of nums1 (s1) and the sum of nums2 (s2).
  2. Calculate difference array: Create a difference array d where d[i] = nums1[i] - nums2[i].
  3. Find maximum subarray sum (Kadane's Algorithm): Apply Kadane's algorithm to the difference array d. This gives the maximum sum of a contiguous subarray in d. This subarray represents the best subarray to swap to maximize the sum of nums1 after the swap. We perform this calculation twice; once for the difference array d (max subarray sum added to s2) and once for -d (max subarray sum added to s1). The max subarray sum for d represents the maximal increase possible for the sum of nums2 (by swapping a segment of nums1 with a segment of nums2), and the max subarray sum for -d represents the maximal increase possible for the sum of nums1 (by swapping a segment of nums2 with a segment of nums1).
  4. Calculate maximum score: The maximum possible score is the maximum of s2 + max_subarray_sum(d) and s1 + max_subarray_sum(-d).

Kadane's Algorithm (for Maximum Subarray Sum):

Kadane's algorithm efficiently finds the maximum sum of a contiguous subarray within a given array. It iterates through the array, keeping track of the current maximum sum (t) and the overall maximum sum (mx). If the current sum t becomes negative, it's reset to 0, as a negative sum would reduce the overall maximum sum.

Time Complexity Analysis:

  • Calculating initial sums: O(n)
  • Creating the difference array: O(n)
  • Kadane's algorithm: O(n) (applied twice)
  • Calculating the maximum score: O(1)

Therefore, the overall time complexity is O(n).

Space Complexity Analysis:

  • The difference array d: O(n)

The overall space complexity is O(n).

Code in Different Languages

The code implementations provided in the original response accurately reflect the algorithm described above. They efficiently compute the maximum score using Kadane's algorithm to find the optimal subarray swap. The functions f(nums1, nums2) within the code implementations represent Kadane's algorithm, which finds the maximum subarray sum of the difference array. Each language's implementation is straightforward and follows the algorithm steps.