Watch 10 video solutions for Maximum Score Of Spliced Array, a hard level problem involving Array, Dynamic Programming. This walkthrough by ThinkCode has 2,374 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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].
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.length1 <= n <= 1051 <= nums1[i], nums2[i] <= 104Problem Overview: You are given two arrays of equal length. You may choose a single subarray and swap it between the arrays. The goal is to maximize the final sum of either array after the splice. The challenge is identifying which segment swap produces the largest increase.
The key observation: swapping a segment does not change both arrays independently. Instead, it transfers the difference between elements in that segment. If you track how much benefit each position gives when swapped, the task reduces to finding a maximum gain subarray.
Approach 1: Dynamic Programming on Difference Array (Kadane) (Time: O(n), Space: O(1))
Compute the base sums of both arrays first. If you swap a segment from nums2 into nums1, the gain at index i becomes nums2[i] - nums1[i]. Build this implicit difference while iterating and apply Kadane's algorithm to find the maximum subarray sum. That value represents the best possible improvement to sum(nums1). Repeat the same process in the opposite direction using nums1[i] - nums2[i] to see if improving nums2 yields a larger result. The final answer is the maximum of both improved sums. The algorithm scans the arrays once and maintains a running maximum gain, so it runs in O(n) time with O(1) space.
This approach works because any valid splice corresponds exactly to a contiguous segment. Kadane's algorithm efficiently identifies the segment where the cumulative difference produces the highest gain.
Approach 2: Greedy Difference Tracking (Time: O(n), Space: O(1))
The same idea can be implemented using a greedy running-gain tracker rather than explicitly describing it as dynamic programming. While iterating through the arrays, compute the difference between elements and maintain a running gain. If the running gain becomes negative, reset it because continuing the segment would only reduce the final score. Track the maximum gain encountered during the scan. Perform this process twice: once assuming you improve nums1 and once assuming you improve nums2. Each pass evaluates whether starting a new splice or extending the current one produces a better result.
This greedy interpretation is essentially Kadane's algorithm but framed as continuously deciding whether to extend the current splice segment or start a new one. It avoids storing additional arrays and only tracks a few running variables.
Conceptually, the problem blends ideas from Array manipulation and maximum subarray problems commonly solved with Dynamic Programming or Greedy strategies.
Recommended for interviews: The Kadane-based dynamic programming approach is what interviewers expect. It shows you can transform the swap operation into a difference-gain problem and apply a classic maximum subarray algorithm. Starting with the naive thought process about swapping segments demonstrates understanding, but recognizing the Kadane reduction shows strong problem-solving skills.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Dynamic Programming (Kadane on Difference) | O(n) | O(1) | Best general solution. Converts splice benefit into a maximum subarray problem. |
| Greedy Difference Tracking | O(n) | O(1) | Same optimal logic implemented with a running gain reset strategy. |
| Bidirectional Gain Evaluation | O(n) | O(1) | Evaluate improvements for both arrays separately to ensure the best final score. |