LeetCode 2321: Maximum Score Of Spliced Array Solution

Master LeetCode problem 2321 (Maximum Score Of Spliced Array), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2321. Maximum Score Of Spliced Array

Hard

Array Dynamic Programming

Problem Explanation

Explanation

To maximize the score, we need to determine the best subarray to swap between the two arrays. We can iterate through all possible subarrays and calculate the score for each swap. The key idea is to find the maximum possible score by choosing the best subarray to swap or not swap at all.

Initialize maxScore to the maximum possible score if no swapping is done.
Iterate through all possible subarrays from left to right.
For each subarray, calculate the potential scores if swapped and update maxScore if a higher score is found.
Return the maxScore.

Time Complexity

The time complexity of this approach is O(n^2), where n is the length of the input arrays. This is because we iterate through all possible subarrays.

Space Complexity

The space complexity is O(1) as we are not using any extra space apart from a few variables.

Solution Code

class Solution {
    public int maxScore(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int maxScore = Math.max(Arrays.stream(nums1).sum(), Arrays.stream(nums2).sum());
        
        for (int left = 0; left < n; left++) {
            for (int right = left; right < n; right++) {
                int tempScore1 = Arrays.stream(nums1).sum() - Arrays.stream(nums1, left, right + 1).sum() +
                                 Arrays.stream(nums2, left, right + 1).sum();
                int tempScore2 = Arrays.stream(nums2).sum() - Arrays.stream(nums2, left, right + 1).sum() +
                                 Arrays.stream(nums1, left, right + 1).sum();
                maxScore = Math.max(maxScore, Math.max(tempScore1, tempScore2));
            }
        }
        
        return maxScore;
    }
}

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